Finding tensions from incline

[gymnast2121]

Blocks 1 and 2 of masses ml and m2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle q with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

I know that the coefficient of friction would be Ff/m1(-9.8)cos(angle).
How would I find the value of the suspended mass M that allows block 1 & 2 to move with constant velocity?

 

[haruspex]

I know that the coefficient of friction would be Ff/m1(-9.8)cos(angle).
How would I find the value of the suspended mass M that allows block 1 & 2 to move with constant velocity?

You won't need to consider the coefficient of friction to answer this part.
Do you understand how to develop free body diagrams when there are multiple interacting bodies? You create unknowns for all of the forces of interaction (the tensions here), draw a free body diagram for each body, and write down the dynamical equations for each body. You also need to make use of kinematic facts, like the fact that some string length does not change. This allows you to deduce that some velocities or accelerations are the same. Where a body has no mass (the pulley here) you know the sum of the forces and sum of torques are zero.

 

[gymnast2121]

So to find the suspended mass M would I create a free body diagram showing all of the forces acting on it then find what equation to find it? Would the Equation involve using sin or cos with the angle?

 

[haruspex]

So to find the suspended mass M would I create a free body diagram showing all of the forces acting on it

That alone won't tell you the mass. You need a free body diagram for each object in the system, and equations obtained from them. When the number of equations matches the number of unknowns, you probably have enough information to solve it.
Since the pulley is massless and frictionless, we can avoid an FBD for that; you just have to figure out how the tension on one side of it relates to the tension on the other side. So three FBDs should do it.
In general, you can get three equations per FBD: two for linear forces along some chosen co-ordinates (often, horizontal and vertical) and one for torque. In the present case there are no torques. So you could get up to 6 equations, but you won't need them all.

What equation do you know relating forces and accelerations in general?

 

[HalfLostAndSearching]

To find the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity, we can use the concept of equilibrium. Since the blocks are moving at a constant velocity, we can assume that the net force acting on them is zero. This means that the forces acting in the horizontal direction must be equal to the forces acting in the vertical direction.

In the horizontal direction, there are two forces acting on block 1: the tension in the string (T) and the frictional force (f). Similarly, there are two forces acting on block 2: the tension in the string (T) and the frictional force (2f). The tension in the string (T) is the same for both blocks since they are connected by the same string.

In the vertical direction, there are three forces acting on block 1: the weight of the block (m1g), the normal force (N) from the inclined plane, and the component of the tension in the string (Tsinθ) acting in the upward direction. Similarly, there are three forces acting on block 2: the weight of the block (m2g), the normal force (N) from the inclined plane, and the component of the tension in the string (Tsinθ) acting in the downward direction.

Since the blocks are moving at a constant velocity, we can equate the horizontal forces and the vertical forces to zero. This gives us the following equations:

Horizontal forces:

T + f = T + 2f
f = 2f

Vertical forces:

m1g + N + Tsinθ = m2g + N + Tsinθ
m1g = m2g

From these equations, we can see that the mass of block 1 (m1) is equal to the mass of block 2 (m2). Therefore, the value of the suspended mass M does not affect the constant velocity of blocks 1 and 2. As long as the masses of blocks 1 and 2 are equal, the system will remain in equilibrium and move at a constant velocity down the inclined plane.