Two rectangular blocks are stacked on a table. The upper block has a mass of m2 and the lower mass is m1 > m2. The coefficient of kinetic friction between the lower block and the table is ųk. The coefficient of static friction between the two blocks is ųk. A string is attached to the lower block, and an external force is applied horizontally, pulling on the string to the right.
without solving the problem, draw appropriate free body diagrams, and write the correct Newton law for the problem.
Determine the maximum force that can be applied to the string without having the upper block sliding off. You have to show the net force in the component directions and consider the acceleration of all objects.
hint: think carefully about what is being accelerated and in which direction the upper block will want to slip.
To determine the maximum force that can be applied to the string without having the upper block sliding off, we need to analyze the forces acting on the system and consider the acceleration of all objects involved.
First, let's draw the free body diagrams for each block:
1. Lower block (m1):
- Weight (mg1) downward (opposite to the upward force exerted by the table).
- Normal force (N1) upward (exerted by the table).
- Force of static friction (fs) rightward (opposite to the impending motion with the upper block).
2. Upper block (m2):
- Weight (mg2) downward (exerted by gravity).
- Force of static friction (fs2) leftward (exerted by the lower block).
Now, let's write the correct Newton's laws for each block:
1. Lower block (m1):
- In the vertical direction:
- N1 - mg1 = 0 (Newton's second law in the vertical direction, since there is no vertical acceleration).
- N1 = mg1 (equilibrium equation).
- In the horizontal direction:
- fs - F = m1 * a (Newton's second law in the horizontal direction).
- fs = μk * N1 (static friction force equation).
- m1 * a = μk * mg1 - F (substituting fs with μk * N1).
2. Upper block (m2):
- In the horizontal direction:
- fs2 = m2 * a (Newton's second law in the horizontal direction).
- fs2 = μs * N2 (static friction force equation, assuming it slides just before the upper block slips).
- m2 * a = μs * mg2 (substituting fs2 with μs * N2).
Now, let's consider the acceleration and direction of the upper block. The upper block will want to slide to the right due to the force applied to the string. Therefore, the direction of acceleration for the upper block is also to the right.
To find the maximum force that can be applied without having the upper block slide off, we need to determine the point at which the static friction force fs between the lower and upper block reaches its maximum value. This occurs just before the upper block starts slipping.
Therefore, we equate the static friction force to the maximum value, which is µs * N1. We substitute this value into the equation for the acceleration of the upper block:
m2 * a = µs * mg2.
Solving this equation will give us the maximum acceleration that can be obtained without slipping. We can then determine the corresponding maximum force using the equation:
F_max = m1 * a + µk * N1.
Finally, substitute the known values of m1, m2, µk, µs, g, and solve for F_max to obtain the answer.