1. I am badly stuck on a few Newton's Laws problems. I hope someone can help me out! Thanks a million!

1) A 5 kg block rests on a 30 degrees incline. The coefficient of static friction between the block and incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding
1a) up the incline?
1b) down the incline?

I really have no clue of solving this. Would 1a and 1b have the same answer?

2) A double incline plane has two boxes of equal masses on opposite sides connected by a rope over a pulley. The left incline is 53 degrees and the right incline is 30 degrees. The coefficient of static and dynamic friction are equal, 0.30 for both boxes.
2a) Show that the system, when released, remains at rest.
2b) If the system is given an initial speed of 0.90m/s to the left, how far will it move before coming to rest?

For 2a, I know that the motion of attempted of the system is to the left beucase the left incline is steeper and the boxes have equal masses. So there are 2 frictional forces, both pointing to the right. Knowing all these, I got that the acceleration is 0.69441m/s^2 [right]. Then, how can the system remain at rest? Can someone explain? I don't understand.

For 2b, v1=-0.90m/s, v2=0, a=0.69441. Using these, I found that the system moves 0.58m to the left before coming to rest. But can I use the acceleration that I got in part a? Is this the correct way of doing it?

2.

3. 1) the answers will be vastly different. Up the incline the horizontal force will partially add to the normal force increasing the friction in addition to the fact that it will be working against gravity. Down the incline, you may have to push in the same direction if the box will slide by itself without any help, other wise the required force in the opposite direction to make slide will be much less because you not only have gravity helping you but your horizontal force will reduce the normal force decreasing the friction.

2) A double incline plane has two boxes of equal masses on opposite sides connected by a rope over a pulley. The left incline is 53 degrees and the right incline is 30 degrees. The coefficient of static and dynamic friction are equal, 0.30 for both boxes.
2a) Show that the system, when released, remains at rest.
This is one of the nastiest problems of this type that I have ever seen.

This next portion in italics is a standard approach to the problem but it doesn't get you anywhere, so you may want to skip over it.

With inclines you aline your coordinate system with x along the incline in the direction of most likely mostion and y perpendicular to the incline.
left box forces
x: mass (9.8m/s^2) sin 53 - Tension - Frictionleft = 0
y: NormalForceLeft - mass (9.8m/s^2) cos 53 = 0
right box forces
x: Tension - mass (9.8m/s^2) sin 30 - Frictionright = 0
y: NormalForceRight - mass(9.8m/s^2) cos 30 = 0

Solve thise for Friction and Normal forces
x: FrictionLeft = mass (9.8m/s^2) sin 53 - Tension
y: NormalForceLeft = mass (9.8m/s) cos 53
right box forces
x: Frictionright = Tension - mass (9.8m/s^2) sin 30
y: NormalForceRight = mass(9.8m/s^2) cos 30

We must show that Friction < .3 Normal Force on both right and left
mass (9.8m/s^2) sin 53 - Tension < .3 mass (9.8m/s^2) cos 53
Tension - mass (9.8m/s^2) sin 30 < .3 mass(9.8m/s^2) cos 30
Or letting w = Tension/mass
we must prove (7.8266m/s^2)- w < (1.7693m/s^2)
and we must prove w - (4.9m/s^2) < (.25981m/s^2)
putting these together means that w must be between 5.1598 and 6.0573m/s^2

Consider the problem with no friction
x: mass a = mass (9.8m/s^2) sin 53 - Tension
right box forces
x: mass a = Tension - mass (9.8m/s^2) sin 30
So a = (9.8m/s^2)(sin 53 - sin 30) = 2.9266m/s^2
and Tension = mass (9.8m/s^2 sin 30 + 2.9266m/s^2)
so Tension = mass 7.8266 m/s^2
(Clearly the frictional forces will reduce this tension)

Try adding the fricion on the 30 degree side only
x: mass (9.8m/s^2) sin 53 - Tension = 0
right box forces
x: Tension - mass (9.8m/s^2) sin 30 - Frictionright = 0
y: NormalForceRight - mass(9.8m/s^2) cos 30 = 0
looks good! Now we get that the Tension = mass 7.8266 m/s^2
So FrictionRight = mass 2.9266 m/s^2
and NormalForceRight = mass 8.487m/s^2
But then .3 NormalForceRight = mass 2.54611m/s^2
Which means that the friction on one side is not enough!

But the friction on the left side can be as much as
.3 NormalForceLeft = .3 mass (9.8m/s) cos 53 = mass 1.7693m/s^2
which is more than enough to make up the difference.

There must be a more elegant way to do this, but I have not thought of it.

4. Wow, how complicated......

5. Ok I have taken another look at this, and I think I understand it better and have cut out the meaningless parts.

First just look at the Normal forces on each of the two boxes
y: NormalForceLeft = mass (9.8m/s^2) cos 53 = mass 5.898 m/s^2
right
y: NormalForceRight = mass(9.8m/s^2) cos 30 = mass 8.487 m/s^2

This gives us the maximum static friction as follows
FrictionLeft = .3 * NormalForceLeft = mass 1.769 m/s^2
FrictionRight = .3 * NormalForceRight = mass 2.546 m/s^2

Consider the problem with no friction
x: mass a = mass (9.8m/s^2) sin 53 - Tension
right box forces
x: mass a = Tension - mass (9.8m/s^2) sin 30
By adding these two equations together we get
a = (9.8m/s^2)(sin 53 - sin 30) = 2.9266m/s^2

This means that the net force on each of the boxes making them move is (mass 2.9266 m/s^2) which is greater than the maximum possible static forces we calculated above. This means that the boxes will move.

The math is the basically same although simplified. My mistake above was in thinking that the we only needed the sum of the frictions on right and left to exceed (mass 2.9266 m/s^2) but it actually needs to exceed twice this because the total mass of the two boxes is 2 times mass.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement