Consider two blocks that are resting one on top of the other. The lower block has mass m2 = 3.5 kg and is resting on a frictionless table. The upper block has mass m1 = 2.7 kg. Suppose the coefficient of static friction between the two blocks is given by μs = 0.5.
Part a) A force of magnitude F is applied as shown in the left figure above. What is the maximum force for which the upper block can be pushed horizontally so that the two blocks move together without slipping?
Part b) A force of magnitude F as shown in the right figure above. What is the maximum force for which the lower block can be pushed horizontally so that the two blocks move together without slipping?
Please Some Help!
To find the maximum force for which the two blocks move together without slipping, we need to analyze the forces acting on the system.
Let's start with part a):
a) To find the maximum force for which the upper block can be pushed horizontally, we need to consider the maximum static friction force between the blocks.
1. Draw a free-body diagram for each block:
- The upper block (m1) has its weight acting downwards (m1 * g) and a force applied horizontally (F) to the right.
- The lower block (m2) has its weight acting downwards (m2 * g) and the static friction force (μs * N) acting to the left.
2. Determine the normal force (N) acting on the lower block:
Since the lower block is resting on a frictionless table, the weight of both blocks is supported solely by the normal force. Therefore, N = (m1 + m2) * g.
3. Write the equations of motion for each block:
For the upper block (m1):
- Net force in the horizontal direction: F - static friction force = m1 * acceleration
- Net force in the vertical direction: N - m1 * g = 0 (since there is no vertical acceleration)
For the lower block (m2):
- Net force in the horizontal direction: static friction force = m2 * acceleration
- Net force in the vertical direction: N - m2 * g = 0 (since there is no vertical acceleration)
4. Determine the maximum static friction force (Ff) between the blocks:
Since the two blocks are on the verge of slipping, the static friction force is at its maximum and can be calculated using the equation Ff = μs * N.
5. Substitute the values into the equations:
For the upper block (m1):
- F - μs * N = m1 * acceleration
For the lower block (m2):
- μs * N = m2 * acceleration
6. We can solve these two simultaneous equations to find the acceleration of the system.
7. Once we have the acceleration, we can determine the maximum force (F) for which the upper block can be pushed. This maximum force is equal to the static friction force, i.e., F = μs * N.
Now let's move on to part b):
b) To find the maximum force for which the lower block can be pushed horizontally, we need to consider the maximum static friction force between the blocks.
1. Draw a free-body diagram for each block:
- The upper block (m1) has its weight acting downwards (m1 * g) and a force applied horizontally (F) to the right.
- The lower block (m2) has its weight acting downwards (m2 * g) and the static friction force (μs * N) acting to the right.
2. Determine the normal force (N) acting on the lower block:
Since the lower block is resting on a frictionless table, the weight of both blocks is supported solely by the normal force. Therefore, N = (m1 + m2) * g.
3. Write the equations of motion for each block:
For the upper block (m1):
- Net force in the horizontal direction: F - static friction force = m1 * acceleration
- Net force in the vertical direction: N - m1 * g = 0 (since there is no vertical acceleration)
For the lower block (m2):
- Net force in the horizontal direction: static friction force = m2 * acceleration
- Net force in the vertical direction: N - m2 * g = 0 (since there is no vertical acceleration)
4. Determine the maximum static friction force (Ff) between the blocks:
Since the two blocks are on the verge of slipping, the static friction force is at its maximum and can be calculated using the equation Ff = μs * N.
5. Substitute the values into the equations:
For the upper block (m1):
- F - μs * N = m1 * acceleration
For the lower block (m2):
- μs * N = m2 * acceleration
6. We can solve these two simultaneous equations to find the acceleration of the system.
7. Once we have the acceleration, we can determine the maximum force (F) for which the lower block can be pushed. This maximum force is equal to the static friction force, i.e., F = μs * N.
By following these steps, you can find the maximum forces for both part a) and part b) of the problem.