# Physics!!

• Jan 15th 2008, 06:22 PM
johett
Physics!!
A 0.5kg wooden block is placed on top of a 1.0kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

thanks in advance for those who helped
• Jan 16th 2008, 01:45 AM
mr fantastic
Quote:

Originally Posted by johett
A 0.5kg wooden block is placed on top of a 1.0kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

thanks in advance for those who helped

I'll take the lower block to be sliding to the right.

Things to note:

1. We want the upper block to be on the point of slipping. If it's not on the point of slipping, then the horizontal force being applied to the lower block is not large enough. We want the maximum force, so the upper block will be just about to start slipping. So the friction force acting on the upper block is $\mu N_2 = 0.35 N_2$. It acts to the RIGHT since the block is on the point of slipping to the left (relative to the bottom block).

2. Both blocks have the same acceleration a because the upper block is not slipping. If it was slipping the acclerations of each block would be different.

3. It always helps to draw a diagram showing all the forces acting on all the objects ....

The 1 kg block has three forces acting on it in the vertical direction: Weight force = (1)(g) = g (down), normal reaction force $N_1$ (up) from the ground, normal reaction force $N_2$ (down) from contact with the bottom of the 0.5 kg mass.

The 1 kg block has two forces acting on it in the horizontal direction: Pulling force P (to the right, say), friction force $\mu N_1 = 0.2 N_1$ since the block is sliding (the force acts to the left).

The 0.5 kg block has two forces acting on it in the vertical direction: Weight force = (0.5)(g) = g/2 (down), normal reaction force $N_2$ (up) from contact with the top of the 1 kg block.

The 0.5 kg block has one force acting on it in the horizontal direction: friction force $\mu N_2 = 0.35 N_2$ since the block is on the point of sliding (this force acts to the RIGHT since the block is on the point of slipping to the left).

4. After considering and doing all of the above you shuold be able to get the following equations:

1 kg block - horizontal direction: $0 = N_1 - N_2 - g$ .... (1)

1 kg block - vertcial direction: $a = P - 0.2 N_1$ .... (2)

0.5 kg block - horizontal direction: $0 = N_2 - g/2$ .... (3)

0.5 kg block - vertical direction: $a/2 = 0.35 N_2$. .... (4)

There are four equations and four unknowns, including P. Solve these four equations simultaneously and you'll get P.
• Jan 16th 2008, 02:10 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
I'll take the lower block to be sliding to the right.

Things to note:

1. We want the upper block to be on the point of slipping. If it's not on the point of slipping, then the horizontal force being applied to the lower block is not large enough. We want the maximum force, so the upper block will be just about to start slipping. So the friction force acting on the upper block is $\mu N_2 = 0.35 N_2$. It acts to the RIGHT since the block is on the point of slipping to the left (relative to the bottom block).

2. Both blocks have the same acceleration a because the upper block is not slipping. If it was slipping the acclerations of each block would be different.

3. It always helps to draw a diagram showing all the forces acting on all the objects ....

The 1 kg block has three forces acting on it in the vertical direction: Weight force = (1)(g) = g (down), normal reaction force $N_1$ (up) from the ground, normal reaction force $N_2$ (down) from contact with the bottom of the 0.5 kg mass.

The 1 kg block has two forces acting on it in the horizontal direction: Pulling force P (to the right, say), friction force $\mu N_1 = 0.2 N_1$ since the block is sliding (the force acts to the left).

The 0.5 kg block has two forces acting on it in the vertical direction: Weight force = (0.5)(g) = g/2 (down), normal reaction force $N_2$ (up) from contact with the top of the 1 kg block.

The 0.5 kg block has one force acting on it in the horizontal direction: friction force $\mu N_2 = 0.35 N_2$ since the block is on the point of sliding (this force acts to the RIGHT since the block is on the point of slipping to the left).

4. After considering and doing all of the above you shuold be able to get the following equations:

1 kg block - horizontal direction: $0 = N_1 - N_2 - g$ .... (1)

1 kg block - vertcial direction: $a = P - 0.2 N_1$ .... (2)

0.5 kg block - horizontal direction: $0 = N_2 - g/2$ .... (3)

0.5 kg block - vertical direction: $a/2 = 0.35 N_2$. .... (4)

There are four equations and four unknowns, including P. Solve these four equations simultaneously and you'll get P.

And here's an easy follow up question for you:

What is the acceleration of the upper block if it's slipping?