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Thread: force vectors of an inclined body

  1. #1 force vectors of an inclined body 
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    Can someone post the formula for the percentages of horizontal vs. vertical force of say a ladder leaning on a wall. Like if the ladder weighs 100 lbs and is inclined at 20 degrees off vertical how much of the 100 is pressing straight down and how much is pressing on the wall. I don't know where my old text book is and maybe it's easier to ask here. Guys at the gym like to brag about leg pressing 600 lbs on the sled at 45 degrees and I like to correct them but they don't like to admit it. At 45 degrees wouldn't that be half of the 450 lbs? But I'd like the formula.


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  3. #2  
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    As far as the weight sled is concerned, we can look at it in terms of energy. Work is force multiplied by distance. This will be equal to the potential energy given to the weight. Potential energy is the weight multiplied by the height it is raised.

    If the weight lifter pushes the 600 lb weight along the sled a distance of 1 foot, then he has not done 600 foot-pounds of work. He has raised it 1 foot times the sine of 45 degrees, which is .707 feet. Therefore he has done 600*.707=424.2 foot pounds of work. Since he pushed it a distance of 1 foot, the force he exerted was 424.2 lb.

    The ladder is a different situation. I'll get back to that later. Gotta go.


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    The ladder problem is a problem in statics, which means nothing is moving. If nothing is moving then the horizontal forces have to add to zero, the vertical forces add to zero, and all moments (i.e., torques) add to zero. If they were otherwise, the system would be accelerating, not static.

    With the ladder, we have to make an assumption about the frictional force of the wall. We will assume it is zero, since the ladder is not designed to grip the wall and the ladder is almost vertical so it isn't pressing very hard against the wall. This means that the vertical force of 100 lb will be borne by the foot of the ladder against the ground.

    Now we need to find the horizontal force. We do this by first calculating moments about the base of the ladder. The weight of the ladder, which is assumed to be concentrated in the middle, is tending to rotate the ladder clockwise, while the force of the wall pressing against the top of the ladder balances this torque in the clockwise direction. Since we assume zero friction, this force is purely horizontal.

    Say the ladder is 10 feet long. If the angle is 20 degrees, then the base of the ladder is 10 sin(20degrees) = 3.42 feet from the wall. The center of gravity of the ladder is half that, or 1.71 feet. So the moment about the base of the ladder is 1.71ft*100lb = 171 lb-ft.
    The wall has a lever arm of 10*cos(20 degrees) = 9.4 feet. The force is the torque divided by the lever arm, so it's 171/9.4=18.2 lb. This means the horizontal (friction) component of the force at the base of the ladder is equal and opposite, 18.2 lb. There is your answer.
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  5. #4  
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    Thanks. I have been working quite high (on the ladder) recently and it makes me wonder about the base slipping and howmuch downward force there is.
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