The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s^2. Disregard any pulley mass or friction in the pulley and let M = 2 kg. The magnitude of the friction force on the block that slides horizontally is 4.6N
What is the tension force between M and 2M?
Description of the blocks location: There is table. In the middle of the table is a block 2M. On the left side of the table is a block M that hang under the left side pulley and on the right side of the table is another Block 2M that hang under the right side pulley.
Draw a free body diagram for M. It is going to accelerate in the direction of the "winning force" at 1.5 m/s^2. Find the net force as it is the vector sum of the forces acting on the block. Use the net force to find the tension.
To find the tension force between M and 2M, we can start by analyzing the forces acting on each block separately.
Let's first consider the block M hanging under the left side pulley. The only forces acting on this block are its weight (mg) and the tension force (T) acting upwards. Since the block is moving with an acceleration of 1.5 m/s^2, we can write the following equation of motion using Newton's second law:
T - mg = ma
Since the mass of the block M is given as M = 2 kg, we can substitute these values into the equation:
T - (2 kg)(9.8 m/s^2) = (2 kg)(1.5 m/s^2)
Simplifying the equation, we get:
T - 19.6 N = 3 N
T = 3 N + 19.6 N
T = 22.6 N
Therefore, the tension force between M and 2M is 22.6 N.
Note: The magnitude of the friction force on the block that slides horizontally is given as 4.6 N. This information is not directly relevant to finding the tension force between M and 2M, but it is still useful to understand the overall scenario.