A large box of mass M is at rest on a horizontal table whose surface is rough; the coefficient of static friction between the table and the block is μ. The box is connected by a taut string to a small block of unknown mass m, initially held horizontally as shown and then released.
1. Assuming the string remains taut and the large box M does not slip on the table, what is the tension in the string just before block m strikes the table? Answer in terms of M, m, g, and μ (mu)
2. Suppose now that the large box begins to slip just before the block strikes the wall. What is the mass of the block? Answer in terms of M, g, and μ (mu).
To answer the given questions, we need to analyze the forces acting on the system and apply the laws of motion. Let's go through each question step by step:
1. To find the tension in the string just before block m strikes the table, we need to consider the forces acting on both blocks. There are two important conditions for this analysis:
a. The large box M does not slip on the table.
b. The string remains taut.
Since the large box does not slip on the table, the frictional force between the box and the table (f_friction) can be given by the equation f_friction = μ * N, where μ is the coefficient of static friction and N is the normal force on the large box M. The normal force N is equal to the weight of the box, which is M * g, where g is the acceleration due to gravity.
Now, let's analyze the forces acting on block m. There are two forces:
a. The tension in the string pulling block m horizontally.
b. The weight of block m, which is m * g.
Since the string remains taut, the tension in the string is equal to the weight of the large box M.
Therefore, just before block m strikes the table, the tension in the string is M * g.
Answer: The tension in the string just before block m strikes the table is M * g.
2. Now, let's consider the scenario where the large box M begins to slip just before the block m strikes the table. This means that the frictional force between the box M and the table has reached its maximum static friction force (f_max). The maximum static friction force can be given by the equation f_max = μ * N, where N is the normal force on the large box M.
Since the maximum static friction force is acting on the large box M, Newton's second law tells us that the force on M is equal to its mass times its acceleration, which is zero since the box M is at rest. Therefore, we have f_max = M * a, where a is the acceleration of the box M.
Now, let's analyze the forces acting on block m. There are two forces:
a. The tension in the string pulling block m horizontally.
b. The weight of block m, which is m * g.
Since the large box M is slipping, the tension in the string is no longer equal to the weight of the large box M. Instead, it is equal to the difference between the maximum static friction force and the weight of the large box M.
Therefore, the tension in the string just before block m strikes the table is f_max - M * g.
Since we are looking for the mass of block m, let's equate the tension in the string to the weight of block m:
f_max - M * g = m * g
Now, we can solve this equation to find the mass of block m:
m = (f_max - M * g) / g
Answer: The mass of block m is (f_max - M * g) / g.
Note: These answers are derived by applying Newton's laws of motion and the equations for friction forces. Actual numerical values can be substituted into these equations to obtain specific results.