A system comprising blocks, a light frictionless pulley, and connecting ropes is shown. The 9 kg block is on a smooth horizontal table The surfaces of the 12 kg block are rough, with μ = 0.30.
In Figure 5.7, the mass M is set so that it descends at constant velocity when released. The mass M is closest to:
In order for the mass M to descend at a constant velocity when released, the forces acting on it must be balanced. This means that the force of tension in the rope connecting mass M to the other blocks is equal to the gravitational force acting on mass M.
The total mass of the system that causes the tension is the sum of the 9 kg and 12 kg blocks, which is 21 kg. Now, the force of tension required to keep the system moving at a constant velocity comes from the friction between the 12 kg block and the surface.
The frictional force can be calculated using the formula: friction = μ * normal force, in this case the normal force is the force from the weight of the 12 kg block, which can be calculated by multiplying the mass (12 kg) by the gravitational acceleration (9.81 m/s²).
Normal force = 12 kg * 9.81 m/s² = 117.72 N
Friction = frictional coefficient * normal force = 0.30 * 117.72 N = 35.316 N
Since the force of tension is equal to the frictional force, we can now solve for the mass M.
Tension = M * g
35.316 N = M * 9.81 m/s²
M = 35.316 N / 9.81 m/s²
M ≈ 3.6 kg
The mass M that descends at a constant velocity when released is closest to 3.6 kg.
To find the mass M, we need to consider the forces acting on the system.
Since the 9 kg block is on a smooth horizontal table, there is no friction acting on it. This means that the only force acting on the 9 kg block is its weight, which can be calculated as:
Weight = mass * gravitational acceleration
Weight = 9 kg * 9.8 m/s^2
Weight = 88.2 N
Now let's consider the forces acting on the 12 kg block. There are two forces acting on it: its weight and the tension in the rope connecting it to the 9 kg block.
The weight of the 12 kg block can be calculated as:
Weight = mass * gravitational acceleration
Weight = 12 kg * 9.8 m/s^2
Weight = 117.6 N
The tension in the rope is the force that is responsible for moving the 12 kg block. It can be calculated using Newton's second law, which states that the net force on an object is equal to the mass of the object times its acceleration:
Net force = mass * acceleration
In this case, the 12 kg block is moving at a constant velocity when released, which means its acceleration is zero. Therefore, the net force on the block is also zero. The tension in the rope is equal in magnitude but opposite in direction to the force of friction acting on the block, which can be calculated as:
Force of friction = coefficient of friction * normal force
The normal force is the force exerted by the table on the 12 kg block, which is equal in magnitude but opposite in direction to the weight of the block:
Normal force = Weight = 117.6 N
Force of friction = 0.30 * 117.6 N
Force of friction = 35.28 N
Since the tension in the rope is equal in magnitude but opposite in direction to the force of friction, we have:
Tension = Force of friction = 35.28 N
Now let's consider the forces acting on the mass M. The force of gravity acting on M is equal to its weight, which is given as:
Weight = mass * gravitational acceleration
Since the mass M descends at a constant velocity, we know the net force acting on it is zero. Therefore, the tension in the rope connected to M must be equal in magnitude but opposite in direction to its weight:
Tension = Weight
Now we can find the mass M by comparing the tension in the rope connected to M with the known weight of the other blocks:
Tension = 35.28 N
Weight of M = Weight of 9 kg block
Since the weight of the 9 kg block is 88.2 N, we can set up the following equation:
Tension = Weight of M
35.28 N = weight of M
35.28 N = mass of M * gravitational acceleration
Solving for the mass of M:
Mass of M = 35.28 N / 9.8 m/s^2
Mass of M = 3.6 kg
Therefore, the mass M is closest to 3.6 kg.
To solve this problem, we need to analyze the forces acting on the system and use Newton's laws of motion.
1. Identify the forces in the system:
- The weight of the 9 kg block (W1 = m1 * g), where g is the acceleration due to gravity.
- The tension in the rope on the 9 kg block (T1).
- The weight of the 12 kg block (W2 = m2 * g).
- The frictional force between the 12 kg block and the table (F_friction = μ * N), where μ is the coefficient of friction and N is the normal force.
2. Apply Newton's first law in the vertical direction:
For the system to descend at a constant velocity, the net force in the vertical direction should be zero.
- T1 - W1 - W2 = 0 (since the 9 kg block and 12 kg block move together)
- T1 = W1 + W2
- T1 = m1 * g + m2 * g
3. Apply Newton's second law for the 12 kg block in the horizontal direction:
The net force in the horizontal direction should be equal to the frictional force.
- T1 - F_friction = 0
- T1 = F_friction = μ * N
- T1 = μ * m2 * g
4. Equate the expressions for T1 obtained from step 2 and step 3:
- m1 * g + m2 * g = μ * m2 * g
- m1 + m2 = μ * m2
- m1 = μ * m2 / (1 - μ)
5. Calculate the value of m1, which is the mass M closest to:
- Substitute the given values: μ = 0.30 and m2 = 12 kg
- m1 = 0.30 * 12 / (1 - 0.30)
- m1 ≈ 0.30 * 12 / 0.70
- m1 ≈ 5.143 kg
Therefore, the mass M, which is closest to the mass of the 9 kg block when released, is approximately 5.143 kg.