Thread: Force on the ground when pulling a weight

1. Hi, I cannnot draw my diagrams yet, but I found this on the web: http://i.stack.imgur.com/U5RB2.jpg , http://i586.photobucket.com/albums/s...pullTorque.png

I am studying forces can you tell me something:
If I pull a weight and exert a horizontal force (they say 300n, but I think a man can easily pull 3000 or more) what is the force acting on the ground throught the feet?
The angle seems 45°, then force is 300, 300*sin y, or 300 /sin y? is this a torque, is it exerted at 45° or horizontally or is it split in components?

Is the equation in the picture right if we consider also gravity?

Thanks a lot

2.

3. All I see is a sketch of a man pulling on a rope. No equations.

If the rope is horizontal and the tension in the rope is 300n, then the frictional force on the feet is also 300n. The total force acting on the ground through the feet will be the vector sum of the man's weight and the horizontal force of 300n. The angle will depend on his weight.

4. Originally Posted by Harold14370
All I see is a sketch of a man pulling on a rope. No equations.

If the rope is horizontal and the tension in the rope is 300n, then the frictional force on the feet is also 300n. The total force acting on the ground through the feet will be the vector sum of the man's weight and the horizontal force of 300n. The angle will depend on his weight.
Thanks if this is visible there is the equation. But I 'd like to exclude gravity.

I'd like to learn what happens when I pull horizontally propping up my feet at 45° on the ground. If I exert a force of 300N what happens at the point of support where feet touch the ground:
- is there a force or a torque?
- the force of 300N is applied obliquely, and the horizontal component is 300*cos y =0.7 =210 N
-or the horizontal component is 300N

Thanks a lot, btw is the equation correct?

5. If you exclude gravity, there is no frictional force, and you can't pull anything at all. Any time you have a force, you can calculate a torque about the point of rotation. Do you know how to calculate a torque? I still don't know what equation you are referring to.

6. Originally Posted by Harold14370
If you exclude gravity, there is no frictional force, and you can't pull anything at all. Any time you have a force, you can calculate a torque about the point of rotation. Do you know how to calculate a torque? I still don't know what equation you are referring to.
I regret the system refuses to accept the image in the post, at the bottom there is an equation that says the torque of g must equal the torque of the pull.

Suppose the feet are blocked to the ground in some way, you brace to the ground and pull , say, at a car, you want to move it a few cm forward.

You are pushing on the ground at 45°, the force you apply to the car is simultaneous applied to the ground (btw what do you call this force?) , torque is F* s, right? like KE. I know that but I am confused as I do not know if I have to consider energy.

But my only concern now is to know how is the force applied to the ground, do we have to split in a parallelogram? the reaction of the ground must be horizontal, right
then is the horizontal component that must be 300 , or the oblique force?

7. Torque and energy have the same units, force multiplied by a length, but don't get confused. They are not the same thing. Energy will not be used in your calculations.

If the man's feet are pinned to the ground, they won't slip, so you don't have to worry about friction. But, the rope will be applying a torque which will tend to rotate the guy in the counterclockwise direction. The man's weight will tend to make him rotate clockwise. You cannot ignore gravity, because that is what is keeping the man from simply being rotated around the pivot point. The reaction of the ground has to oppose both the weight which is vertical and the pull of the rope which is horizontal, so you are wrong to think it is only horizontal.

This is a problem in statics. In a static system where there is no acceleration involved, all forces add up to zero and all torques add up to zero. You attack it in parts. Do the forces first, then do the torques.

8. Originally Posted by Harold14370
This is a problem in statics. In a static system where there is no acceleration involved, all forces add up to zero and all torques add up to zero. You attack it in parts. Do the forces first, then do the torques.
You are right Harold, that would be too much for me at the moment. That's why I said : do not consider gravity for now. suppose the legs make a large base so that the centre of gravity is inside and the force is cancelled by the reaction of the ground, as in the first picture.

Now let's consider only the force on the rope and not the torque. I need to learn this basic manner to deal with forces. I read that the parallelogram of vectors is useful also for forces. So , I thought, the 300N pull on rope is applied at 45° with the ground, so it has to be split( is this the right term) in two vectors (right term?) each 300N*cos y (as sin=cos), the vertical one is nullified on the ground (same as g) so in conclusion the horizontal pull on the posite x-axis is the force that must be opposed and blocked by static friction, like a staring block or a hole in the ground.

Please correct what is wrong and repeat this description of the force the pull on the rope exerts on the ground with appropriate and correc technical terms

9. Is the 300n the total force you are excerting or is it the force that is applied to the weight?

10. Originally Posted by Sealeaf
Is the 300n the total force you are excerting or is it the force that is applied to the weight?
I have been told that force on the rope/weight is total force, isn't it so?

11. http://s47.photobucket.com/user/lisa...g.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
Originally Posted by Harold14370
Torque and energy have the same units, force multiplied by a length, but don't get confused. They are not the same thing. Energy will not be used in your calculations.

If the man's feet are pinned to the ground, they won't slip, so you don't have to worry about friction. But, the rope will be applying a torque which will tend to rotate the guy in the counterclockwise direction. The man's weight will tend to make him rotate clockwise. You cannot ignore gravity, because that is what is keeping the man from simply being rotated around the pivot point. The reaction of the ground has to oppose both the weight which is vertical and the pull of the rope which is horizontal, so you are wrong to think it is only horizontal.

This is a problem in statics. In a static system where there is no acceleration involved, all forces add up to zero and all torques add up to zero. You attack it in parts. Do the forces first, then do the torques.
I cannot insert this image, can someone fix it?can you tell me why direct upload has faild? its jpeg alright
http://s47.photobucket.com/user/lisa...]=1&sort=1&o=0
http://s47.photobucket.com/user/lisa...]=1&sort=1&o=0
http://s47.photobucket.com/user/lisa...]=1&sort=1&o=0

http://s47.photobucket.com/user/lisa...]=1&sort=1&o=0

I have copied it more times because after you click once, it doesn't work anymore, it's odd, my fault?

Hi harold can you tell me if the forces are correctly represented, friction is only along the x-axis?

12.

13. I find this diagram confusing. It does seem to be showing that there is a force of 300N on the leg at a 45 degree angle to horizontal, which has a horizontal component of 210 (the friction) and a vertical component of 210 (man's weight). However this cannot be in balance with the 300n tension in the rope which is horizontal. The 300N force on the rope cannot be applied at a 45 degree angle. Ropes are floppy things and they can only apply a force directly in line with the rope.
In order for the forces to be balanced, the sum of all horizontal forces must be zero, so if the friction is 210 it cannot balance a tension of 300 in the rope. I think we need to have a complete statement of the problem. Something is missing.

14. Originally Posted by Harold14370
there is a force of 300N on the leg at a 45 degree angle to horizontal, which has a horizontal component of 210 (the friction) and a vertical component of 210 (man's weight).

However this cannot be in balance with the 300n tension in the rope which is horizontal. The 300N force on the rope cannot be applied at a 45 degree angle. Ropes are floppy things and they can only apply a force directly in line with the rope.
In order for the forces to be balanced, the sum of all horizontal forces must be zero, so if the friction is 210 it cannot balance a tension of 300 in the rope. I think we need to have a complete statement of the problem. Something is missing.
Originally Posted by Harold14370
Man's weight is in green and is 980N is completely applied to the ground.

The pull on the rope is 300 N and is applied from CM to the foremost foot 300N
As the force is applied at an angle is split into 2 components, the horizontal would make him slip if there is no friction, if its value is > 210 the man will not slip.

If the man is not pulling he is in equilibrium, if the resistance of the truck is < 300N he will lean back a few cm .

Does this Resolution of Forces apply here?
what is wrong? I'll make a new sketch . why can't I upload here.

Thanks a lot.

15. The procedure I used for the image was as follows. Click on the insert image icon. Click on the "from url" tab. Paste url. Uncheck the box that says "retrieve remote file and reference locally." Click OK.

It still isn't making sense to me, for the same reason I said before. You can't get a 300N horizontal force from a 300N force applied at a 45 degree angle.

16. Originally Posted by Harold14370
The procedure I used for the image was as follows. Click on the insert image icon. Click on the "from url" tab. Paste url. Uncheck the box that says "retrieve remote file and reference locally." Click OK.

It still isn't making sense to me, for the same reason I said before. You can't get a 300N horizontal force from a 300N force applied at a 45 degree angle.
I made a new sketch, hope it's clearer.
=141333040&filters[recent]=1&sort=1&o=0]PullT_zps6c8c5be2.jpg Photo by lisa0rg | Photobucket
I followed your instructions, here is the result.

Ok, if it is wrong tell me what is there to be changed and why we must not apply the parallelogram rule here. I put 290 for the reaction of the truck as I wan to know what happens when the truck is budging.

Thanks, you are very kind

17. Not sure what problem you're having but here is your other sketch. Still not right. You are showing the man's whole weight of 980 on the back foot, but some of it will be on the front foot. The back leg is vertical, so it cannot be providing any horizontal force. The compression force on the front leg is going to have to be bigger to have a 300N horizontal component. It has to be 300*cos(45 degrees) in order to have a 300N horizontal component.

18. Originally Posted by Harold14370

Not sure what problem you're having but here is your other sketch. Still not right. You are showing the man's whole weight of 980 on the back foot, but some of it will be on the front foot. The back leg is vertical, so it cannot be providing any horizontal force. The compression force on the front leg is going to have to be bigger to have a 300N horizontal component. It has to be 300*cos(45 degrees) in order to have a 300N horizontal component.
Wait shouldn't the compression force on the leg be 300*tan(45)? Put another way, since the angle is 45 degrees, the horizontal and vertical components should be equal. This puts the compression on the leg at ~424N. Thus the sideways push the front foot applies to the ground is 300N, the downward force is 300N , and this leaves 680N to be supported by the back foot.

19. Originally Posted by Janus
Originally Posted by Harold14370

Not sure what problem you're having but here is your other sketch. Still not right. You are showing the man's whole weight of 980 on the back foot, but some of it will be on the front foot. The back leg is vertical, so it cannot be providing any horizontal force. The compression force on the front leg is going to have to be bigger to have a 300N horizontal component. It has to be 300*cos(45 degrees) in order to have a 300N horizontal component.
Wait shouldn't the compression force on the leg be 300*tan(45)? Put another way, since the angle is 45 degrees, the horizontal and vertical components should be equal. This puts the compression on the leg at ~424N. Thus the sideways push the front foot applies to the ground is 300N, the downward force is 300N , and this leaves 680N to be supported by the back foot.
I am sorry I am not able to upload, Redpanda said it's because I am new.

I do not understand your remarks, Janus. Gravity is applied to CoM and follows a vertical line, if it happens to fall on the backfoot that means only that he can lift is right foot and still be balanced, Is this wrong?.

The man is exerting a F of 300N propping on his right foot, as long as the truck doesn't move, he can lift/double bend is left leg or stretch it to join the right foot, and still will not fall. All F is therefore applied to the right foot at 45°. I only applied the rules I found on the composition/ resolution of forces: the parallelogram.

All the forces are balanced and so 3rd law is respected, if the reaction of the truck is 290 < 300 then the truck will budge a few centimetres and the rope/man accelerates and 2nd law is also respected.
The block before the foot provides a unlimited friction, so can easily take 210 N.

If the numbers are wrong, please tell where to find the relative law and the explanations.

Thanks, you are all exquisite in this forum. Congrats

20. Originally Posted by Janus
Originally Posted by Harold14370

Not sure what problem you're having but here is your other sketch. Still not right. You are showing the man's whole weight of 980 on the back foot, but some of it will be on the front foot. The back leg is vertical, so it cannot be providing any horizontal force. The compression force on the front leg is going to have to be bigger to have a 300N horizontal component. It has to be 300*cos(45 degrees) in order to have a 300N horizontal component.
Wait shouldn't the compression force on the leg be 300*tan(45)? Put another way, since the angle is 45 degrees, the horizontal and vertical components should be equal. This puts the compression on the leg at ~424N. Thus the sideways push the front foot applies to the ground is 300N, the downward force is 300N , and this leaves 680N to be supported by the back foot.
Oops I should have said the compression on the front leg is 300/cos 45 deg instead of 300*cos 45. That's the hypotenuse of the triangle with legs of 300N.

21. Not sure what problem you're having but here is your other sketch. Still not right. You are showing the man's whole weight of 980 on the back foot, but some of it will be on the front foot. The back leg is vertical, so it cannot be providing any horizontal force. The compression force on the front leg is going to have to be bigger to have a 300N horizontal component. It has to be 300*cos(45 degrees) in order to have a 300N horizontal component.
Wait shouldn't the compression force on the leg be 300*tan(45)? Put another way, since the angle is 45 degrees, the horizontal and vertical components should be equal. This puts the compression on the leg at ~424N. Thus the sideways push the front foot applies to the ground is 300N, the downward force is 300N , and this leaves 680N to be supported by the back foot.[/QUOTE]
Oops I should have said the compression on the front leg is 300/cos 45 deg instead of 300*cos 45. That's the hypotenuse of the triangle with legs of 300N.[/QUOTE]
I think I understand what you mean, but the F from the pull on the forefoot cannot exceed 300 N, so we must add another 90 N on the x-axis from gravity, which makes 127 at 45 and on the backfoot are left 853, which is also more natural as Janus remarked.
Can you think of a rough but quick way to determine how wight is distributed on the two legs?

Thanks, you all have been so kind,
please tell me the right terms:
can I say the pull on the foot is SPLIT in two vectors/components according to the parallelogram?
I usually say that the force applied on the truck/rope is 'discharged' on the ground via the foot
Can you re-frame it in a correct manner?

Thanks

I think I understand what you mean, but the F from the pull on the forefoot cannot exceed 300 N, so we must add another 90 N on the x-axis from gravity, which makes 127 at 45 and on the backfoot are left 853, which is also more natural as Janus remarked.
Can you think of a rough but quick way to determine how wight is distributed on the two legs?
Janus gave you the right numbers. You don't have to add another 90N. The pull on the forefoot cannot exceed 300N in the horizontal direction, but the total force on the foot does exceed the tension in the rope. It has to, because only the horizontal component of this force opposes the tension in the rope.
There is no shortcut. The process is always the same. You add up all the forces and set the sum to zero. In this case you have a force on the man in the x direction of -300N and a force in the y direction of -980. These must be opposed by the sum of the reaction forces transmitted by the ground through the man's feet. We know that the man's back foot will not exert any force in the x direction, so the x component of the force on the front foot is equal in magnitude to the rope tension. We also know that this force on the front foot is applied at a 45 degree angle, so the x component of the force on the front foot is the same as the y component. This plus the y component of the force on the back foot must add up to 980.
Thanks, you all have been so kind,
please tell me the right terms:
can I say the pull on the foot is SPLIT in two vectors/components according to the parallelogram?
I usually say that the force applied on the truck/rope is 'discharged' on the ground via the foot
Can you re-frame it in a correct manner?

Thanks
I would not say it is split. I would say it is the vector sum of the two forces.
I would say that the force applied on the truck/rope is transmitted to the ground via the foot.

23. Originally Posted by Harold14370
I would not say it is split. I would say it is the vector sum of the two forces.
I would say that the force applied on the truck/rope is transmitted to the ground via the foot.
Thanks, Harold,
I have corrected my sketch, can you tell me how to put signs? is minus what is toward -x or what is on the man? Should all red arrows be minus? but the truck reaction points away from the man.

As to gravity, how do I calculate the distribution on the legs, It seems I cannot apply the parallelogram.
Suppose the CM of the man is 1m from the forefoot at 45° , so at 71 cm from the ground and the backfoot is at 15 ° on the other side of the normal from CM, shall I use pytagoras or what. if gravity is 124 N on the forefoot diagonally on 1m, what is g on the backfoot laying 19cm behind the normal, on a length of 73.2?

24. Is this your new sketch? You are still not getting the idea. It looks like you got the 424N force right but I don't know where you are getting the R212, G88, or G124. Also you are still showing a 300N diagonal force. What is that?

25. If 424 is right , it is the sum of the push on foot needed to pull 300N plus 124 from gravity, 980 is distributed unequally between right foot 12a and left foot 856
I have completed the sketch
ManPullsT_zps61720dbc.jpg Photo by lisa0rg | Photobucket

Can you tell me how you take the image into the post?
Is the distribution of gravity correct? is there a formula?

Thanks

26. That looks better.
I explained how to post images before. If that's not working for you, I guess it could be because of your post count. I'm not sure how many posts you need before you can post images.

27. So is this distribution of forces correct at last? There was one vector missing ,yet. I fixed it

=141333040&filters[recent]=1&sort=1&o=1]ManPullsTr_zps20248e3a.jpg Photo by lisa0rg | Photobucket

Image is rejected both from my PC and from URL.
Let me know it that diagram is perfect as it is, I used minus for -x an -y axis is that correct?

AS to gravity I put numbers of fantasy, can you show me how to find the exact weight on the back leg here and in general?parallelogram, vector addition or what?

=141333040&filters[recent]=1&sort=1&o=0]manastride_zpscc5a2c16.jpg Photo by lisa0rg | Photobucket

Thanks, Harold

28. Wait a minute. You still have the weight on the back foot wrong. It's 980 (total weight)-300 (on the front foot) = 680, like Janus said.

29. Originally Posted by Harold14370
Wait a minute. You still have the weight on the back foot wrong. It's 980 (total weight)-300 (on the front foot) = 680, like Janus said.
On the front foot g is 124, 980-124 = 856, then you have to add the vertical component of the foreleg whic is 88. The total, of course will not sum up to 980, because the force has been 'split' twice.

30. I have no idea where you are getting numbers like 124 or 88. What are those?

31. Originally Posted by Harold14370
I have no idea where you are getting numbers like 124 or 88. What are those?
I have made some sketches startingfrom the simplest case:

-1)If one foot is under CM you can transmit all your weight on your left leg if no other force is acting on you, you can lift your right foot from the ground and then adjust your posture so that CM is aligned with the center of the left foot. gravity A-B
Position B, one foot bang under CM:
Position B
If the rear foot is at B the whole weight is on the LF, and no g is acting on RF, you may lift it and nothing happens.But, if it is beyond B (at A, or further) the weight is shared in growing proportions by RF.
When you are pulling the rope, the appropriate position of CM/lF is found authomatically, pushing against the block at the required force. Here are the forces necessary to compensate a pull of 300 N:

-2) Position A
As you can see there is a horizontal F-x on RF (-88N) which is opposed by the Block pinned to the ground (do you call this 'friction'?) and a positive Fx from gravity on LF (DaleSpam disagrees but any civil engineer will confirm that a weight on sloping supports (and the legs are sloping here) generates a horizontal normal push/ force). In the sketch it's 148N but it can vary and adjust itself to needs) .This x-force will anyway compensate the -x force you supposed will produced when you pull on the rope.

I figured an angle of 10°-15° for LF, but I am not sure when 148 is right. What we know for sure is that the R-leg angle must be 45°,

If g on LF 856N, the left foot will gain the necessary 124 N missing to reach the balance ([124*cos45° =] -88-212) the horizontal F-x which is necessary to compensate the torque fom the rope +300N.

The force coming from the pull on the rope (+300) cannot exceed -300, but in order to get a horizontal -x force = -300 , the total force on the R-leg must be 424,
the -124N missing are provided by the pull of g on RL , if the rear leg are getting only -856N.
Is it clear now?

32.

Originally Posted by Harold14370
I have no idea where you are getting numbers like 124 or 88. What are those?
I have made some sketches startingfrom the simplest case:

-1)If one foot is under CM you can transmit all your weight on your left leg if no other force is acting on you, you can lift your right foot from the ground and then adjust your posture so that CM is aligned with the center of the left foot. gravity A-B
Position B, one foot bang under CM:
Position B
If the rear foot is at B the whole weight is on the LF, and no g is acting on RF, you may lift it and nothing happens.But, if it is beyond B (at A, or further) the weight is shared in growing proportions by RF.
When you are pulling the rope, the appropriate position of CM/lF is found authomatically, pushing against the block at the required force. Here are the forces necessary to compensate a pull of 300 N:

-2) Position A
As you can see there is a horizontal F-x on RF (-88N) which is opposed by the Block pinned to the ground (do you call this 'friction'?)
No, I can't see that. Why is it 88? If the tension on the rope is 0, it should be 0.
and a positive Fx from gravity on LF (DaleSpam disagrees but any civil engineer will confirm that a weight on sloping supports (and the legs are sloping here) generates a horizontal normal push/ force). In the sketch it's 148N but it can vary and adjust itself to needs) .This x-force will anyway compensate the -x force you supposed will produced when you pull on the rope.
So you just made up the 148? Why?
I figured an angle of 10°-15° for LF, but I am not sure when 148 is right. What we know for sure is that the R-leg angle must be 45°,
Why does this angle have to be 45 degrees? Is this an assumption in your problem or what is it?

If g on LF 856N, the left foot will gain the necessary 124 N missing to reach the balance ([124*cos45° =] -88-212) the horizontal F-x which is necessary to compensate the torque fom the rope +300N.

The force coming from the pull on the rope (+300) cannot exceed -300, but in order to get a horizontal -x force = -300 , the total force on the R-leg must be 424,
the -124N missing are provided by the pull of g on RL , if the rear leg are getting only -856N.
Is it clear now?
Clear as mud. There is no missing 124N. I think you are trying to add scalars to vectors. That doesn't work.

34. Originally Posted by Harold14370
Is it clear now?
Clear as mud. There is no missing 124N. I think you are trying to add scalars to vectors. That doesn't work.[/QUOTE]
Of course in this diagram is not seen why we need 124N, this explains where that force comes from.
When you pull on the rope ( post #25) you get 300N and you need a friction of -300 to compensate. The pull from the rope gove -300N at 45° on the leg, which is only -212 on the x axis. the missing -88N are provided by gravity on the fore leg: -124 which (*cos45°) becomes -88.
Is that OK now?

Of course in this diagram is not seen why we need 124N, this explains where that force comes from.
When you pull on the rope ( post #25) you get 300N and you need a friction of -300 to compensate. The pull from the rope gove -300N at 45° on the leg, which is only -212 on the x axis.
How did you determine 212 on the x axis? The horizontal pull on the rope of 300N requires 300/cos(45deg)=424N applied at an angle of 45 degrees. The vertical component of this at 45 degrees is 300N. There is no 212 involved.
the missing -88N are provided by gravity on the fore leg: -124 which (*cos45°) becomes -88.
Is that OK now?
No. There is no missing 88N.

36. Originally Posted by Harold14370
Of course in this diagram is not seen why we need 124N, this explains where that force comes from.
When you pull on the rope ( post #25) you get 300N and you need a friction of -300 to compensate. The pull from the rope gove -300N at 45° on the leg, which is only -212 on the x axis.
How did you determine 212 on the x axis? The horizontal pull on the rope of 300N requires 300/cos(45deg)=424N applied at an angle of 45 degrees. The vertical component of this at 45 degrees is 300N. There is no 212 involved.
the missing -88N are provided by gravity on the fore leg: -124 which (*cos45°) becomes -88.
Is that OK now?
No. There is no missing 88N.
Yes the horizontal pull requires -424N on the leg but by itself it can only provide 300, I supposed.

Is not so? If you are standing up and try to pull a pole out of the ground, do your legs exert same or greater force on the ground? can you exert 300N on the pole and 424N on the ground? if it is so please explain how it works, and where the extra energy comes from.

If it is so then there are 124 N missing on the left leg, and these are provided by g, I supposed.

Originally Posted by Harold14370
Of course in this diagram is not seen why we need 124N, this explains where that force comes from.
When you pull on the rope ( post #25) you get 300N and you need a friction of -300 to compensate. The pull from the rope gove -300N at 45° on the leg, which is only -212 on the x axis.
How did you determine 212 on the x axis? The horizontal pull on the rope of 300N requires 300/cos(45deg)=424N applied at an angle of 45 degrees. The vertical component of this at 45 degrees is 300N. There is no 212 involved.
the missing -88N are provided by gravity on the fore leg: -124 which (*cos45°) becomes -88.
Is that OK now?
No. There is no missing 88N.
Yes the horizontal pull requires -424N on the leg but by itself it can only provide 300, I supposed.

Is not so? If you are standing up and try to pull a pole out of the ground, do your legs exert same or greater force on the ground? can you exert 300N on the pole and 424N on the ground? if it is so please explain how it works, and where the extra energy comes from.

If it is so then there are 124 N missing on the left leg, and these are provided by g, I supposed.
Let's simplify things to try and clear it up. Get rid of the man. replace him with a straight rod, anchored to the ground but free to pivot on that point. The rod is leaning at an angle of 45 degrees. the horizontal rope is tied to it and exerts 300N horizontally. Hanging from the same point is a 300N weight. At this point everything is in equilibrium. The 300N downward on the rod exactly balances out the 300 N horizontally. So what is the force acting along the length of the rod? it is the vector sum of the two forces acting on the end. Complete the force parallelogram for these two forces and you get a diagonal force of 424N.

Now increase the hanging weight to 980N, this is 680N greater than what it takes to balance the 300N horizontal force and things are no longer in equilibrium. So we add a second rod under the weight to support the excess (this takes the roll of the man's rear leg), and the downward force it feels is 680 N.

Yes the horizontal pull requires -424N on the leg but by itself it can only provide 300, I supposed.
Are you familiar with the concept of mechanical advantage? You can indeed apply a greater force than you are pulling with, using a lever.
Is not so? If you are standing up and try to pull a pole out of the ground, do your legs exert same or greater force on the ground? can you exert 300N on the pole and 424N on the ground? if it is so please explain how it works, and where the extra energy comes from.
There is no energy involved in the calculation. This is a problem in statics, so we are just comparing forces. .

39. Originally Posted by Janus
. So what is the force acting along the length of the rod? it is the vector sum of the two forces acting on the end. Complete the force parallelogram for these two forces and you get a diagonal force of 424N.

.
- I read that here what matters are torques, not forces. If it is so the torques act tangentially and not verically, this picture is like a pendulum isn't it?

- If , on the contrary I consider perpendicular forces and use the parallelogram, I do not get a square anymore, but a rectangle 2.5 * 1, therefore the resulting force is not at 45 ° but at 68°. How do I calculate the force acting on the pivot at 45°?

Thanks to you and Harold

40.

41. As you have depicted it, the force on the pivot point where you have "F on P?" is 424N, just as we calculated it before. The heavy red arrow which you have labeled 450 should be 300N if you are trying to show the downward force needed to counteract the rotation produced by the 300N pull on the rope. If you want to work it out using torques, you would do it as follows. The torque produced by the tension in the rope is counterclockwise. The force applied is at distance of 1 meter *sin (45) = .707 meters above the pivot point, so the counterclockwise torque is 300*.707=212 newton-meters. The clockwise torque is provided by a vertical force which is applied at a distance of 1 meter * cos(45) = .707 meters to the right of the pivot point. So if the clockwise torque equals the counterclockwise torque, it has to be equal to 212 newton-meters/.707 meters = 300N.

42. Originally Posted by Harold14370
As you have depicted it, the force on the pivot point where you have "F on P?" is 424N, just as we calculated it before. The heavy red arrow which you have labeled 450 should be 300N if you are trying to show the downward force needed to counteract the rotation produced by the 300N pull on the rope.

If you want to work it out using torques, you would do it as follows. The torque produced by the tension in the rope is counterclockwise. The force applied is at distance of 1 meter *sin (45) = .707 meters above the pivot point, so the counterclockwise torque is 300*.707=212 newton-meters. The clockwise torque is provided by a vertical force which is applied at a distance of 1 meter * cos(45) = .707 meters to the right of the pivot point. So if the clockwise torque equals the counterclockwise torque, it has to be equal to 212 newton-meters/.707 meters = 300N.
Thanks Harold, the case of 300 and 300 is simple, and I understood it, can you give me the values when g is 750?
as I said, the resulting force from the parallelogram is at 68° and measures 807.8, but what is the value on the pivot at 45°?
can you correct the signs if they are wrong?

Thanks a lot

43. I'm not sure I understand the question. Could you state the actual problem you are trying to solve? If the force of gravity is 750 and the angle is 45 degrees, then the forces no longer balance and the load accelerates. It is no longer a statics problem, but a dynamics problem.

44. Originally Posted by Harold14370
I'm not sure I understand the question. Could you state the actual problem you are trying to solve? If the force of gravity is 750 and the angle is 45 degrees, then the forces no longer balance and the load accelerates. It is no longer a statics problem, but a dynamics problem.
Before the load accelerates, what is the force actng on the pivot resulting from 300 horizontal and 750 vertical? is it possible to find that value? and the clockwise torque is correct at 530? and the balance is really a colockwise torque of 318?
Are the signs correct?

Originally Posted by Harold14370
I'm not sure I understand the question. Could you state the actual problem you are trying to solve? If the force of gravity is 750 and the angle is 45 degrees, then the forces no longer balance and the load accelerates. It is no longer a statics problem, but a dynamics problem.
Before the load accelerates, what is the force actng on the pivot resulting from 300 horizontal and 750 vertical? is it possible to find that value? and the clockwise torque is correct at 530? and the balance is really a colockwise torque of 318?
Are the signs correct?
There is no "before the load accelerates." You cannot have a force without an equal and opposite opposing force, according to Newton's laws.

46. Originally Posted by Harold14370
There is no "before the load accelerates." You cannot have a force without an equal and opposite opposing force, according to Newton's laws.
Must g have an opposite force? Suppose that rod and load is a pendulum rised by 135° , does gravity has an opposite force there? what would be the force on the pivot if the load were ca.75kg?

47. You shouldn't call it g if you mean the weight. "g" is the symbol used for acceleration due to gravity.

Is the pendulum swinging? Then again it is a problem in dynamics, not statics, and you have masses that are accelerating. Please be precise in your statement of the problem.

48. Originally Posted by Janus
Let's simplify ....diagonal force of 424N.
Now increase the hanging weight to 980N, this is 680N greater than what it takes to balance the 300N horizontal force and things are no longer in equilibrium. So we add a second rod under the weight to support the excess (this takes the roll of the man's rear leg), and the downward force it feels is 680 N.
Thanks Janus, I have done what you said:
=141333040&filters[recent]=1&sort=1&o=0]T300_zps991ca03b.jpg Photo by lisa0rg | Photobucket

Let's change the weight to 750, it's easier. So 300N are cancelled out by the rope, to get equilibrium we need to find where, at what distance from CM, at what angle the rear foot must be in order to carry 400N not 1N more, nor less. I showed in post #31 the rear foot behind the normal line, but I realize this is wrong and the foot must be before Position B, because the normal from CM must fall outside the body in order to exercise a force of 300N.

I started a new thread, but it seems to complex for me, can you tell me straight where exactly I must place the second rod?

Thanks.

Thanks Janus, I have done what you said:
=141333040&filters[recent]=1&sort=1&o=0]T300_zps991ca03b.jpg Photo by lisa0rg | Photobucket

Let's change the weight to 750, it's easier. So 300N are cancelled out by the rope, to get equilibrium we need to find where, at what distance from CM, at what angle the rear foot must be in order to carry 400N not 1N more, nor less.
If the weight is 750 and 300 of this is cancelled by the rope, the rear foot has to carry 450. There is no angle you can put it where it only carries 400.
I showed in post #31 the rear foot behind the normal line, but I realize this is wrong and the foot must be before Position B, because the normal from CM must fall outside the body in order to exercise a force of 300N.
Not sure what you are saying. The rear foot can be directly below the center or it can be at an angle. The further it is off vertical the stronger it has to be.

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