A 4.0 kg block is put on top of a 5.0 kg block. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 16 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together and
To find the maximum horizontal force F that can be applied to the lower block so that the blocks will move together, we need to analyze the forces acting on the system.
First, let's draw a free-body diagram for each block:
For the top block (4.0 kg):
- There is a weight acting downward with a magnitude of (4.0 kg) * (9.8 m/s^2) = 39.2 N.
- There is a normal force exerted by the bottom block that cancels out part of the weight, since the blocks are in contact.
- There is a frictional force acting horizontally, opposing the applied force.
For the bottom block (5.0 kg):
- There is a weight acting downward with a magnitude of (5.0 kg) * (9.8 m/s^2) = 49.0 N.
- There is a normal force exerted by the table that cancels out part of the weight, since the block is in contact with the table.
- There is a frictional force acting horizontally, opposing the relative motion between the blocks.
Since the table is frictionless, the frictional force between the bottom block and the table is zero.
To determine the maximum force F, we need to consider two cases:
1. When the blocks are at rest:
In this case, the force of friction between the blocks is at its maximum, equal to the product of the coefficient of static friction (μs) and the normal force (Fn).
2. When the blocks are moving together:
Once the blocks start moving, the force of friction between the blocks changes and becomes the kinetic friction force, which is equal to the product of the coefficient of kinetic friction (μk) and the normal force (Fn).
Given that a horizontal force of at least 16 N is required to make the top block slip on the bottom one, we can assume that the maximum static friction force is 16 N.
Using Newton's second law, we can set up the equation for each block:
For the top block:
ΣF_top = m_top * a = F_applied - F_friction_top
For the bottom block:
ΣF_bottom = m_bottom * a = -F_friction_bottom
Since the blocks are moving together, their accelerations are equal, so we can set the equations equal to each other:
m_top * a = m_bottom * a
Simplifying and substituting the known values:
4.0 kg * a = 5.0 kg * a
The mass cancels out, so we have:
4.0 = 5.0
This equation is not possible, which means that at the maximum force F, the blocks will not move together. Therefore, we cannot determine the maximum force F that can be applied to the lower block to make the blocks move together on a frictionless table.