Rigid Bodies and Free Body Diagrams (w pulley)

[moo5003]

Question: "A block of mass m1 = 1.87 kg and a block of mass m2 = 5.84 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.180m and mass M = 12.7 kg. These blocks are allowed to move on a fixed block-wedge of angle 31.4 degrees as in the figure. The coefficient of kinetic friction is .357 for both blocks.

A) Using free-body diagrams of both blocks and of the pulley, determine the acceleration of the two blocks.

B) Determine the tensions in the string on both sides of the pulley.
"
Diagram

BLOCK(m1) Pulley
__________________\
\\
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\\\\\BLOCK (m2)
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ANGLE\\\\
_________________________\\

My basic question here is how I factor in the pulley. I'm assuming I need to incorporate torques with the pulley and convert them to forces such that I can find a F(net)=ma for each block.

Block m1 WORK DONE:
X: F(net) = T1(Tension of string) - F(friction)
Y: F(net) = -m1g + N = 0

Block m2 WORK DONE:
X: F(net) = m2gsin(Theta) - T2(String) - F(friction)
Note: X is directed along the slope of the ramp.
Y: F(net) = -m2gcos(Theta) + N = 0
Perpindicular to ramp.

Thats as far as I got. Any help is appreciated.

 

[Doc Al]

So far, so good. Write F(friction) in terms of the normal force and the coefficient of friction.

Treat the pulley as a rotating disk with the two tensions exerting torques on it.

Now apply Newton's 2nd law to all three objects. Be sure to incorporate the system constraint relating the motion of all three: Since they are attached by a string, the linear acceleration of both masses must be the same (in magnitude). What's the relationship between the linear acceleration of the masses and the angular acceleration of the pulley?

 

[quicknote]

I would approach this problem by first defining the system and identifying all the forces acting on it. In this case, the system consists of the two blocks, the pulley, and the string connecting them. The forces acting on the system are: the weight of both blocks (mg), the normal force from the wedge (N), the tension in the string (T1 and T2), and the friction force (μmg) acting in the opposite direction of motion.

To determine the acceleration of the blocks, we can use Newton's Second Law (F=ma) for each block separately. For block m1, the forces acting on it are the tension in the string (T1) and the friction force (μmg). Using the free body diagram, we can write the equation:

m1a = T1 - μm1g

Similarly, for block m2, the forces acting on it are the weight of the block (m2g), the normal force (N), the tension in the string (T2), and the friction force (μm2g). Using the free body diagram, we can write the equation:

m2a = m2gsin(θ) - T2 - μm2gcos(θ)

We also need to consider the rotation of the pulley. Since the pulley is a rigid body, it will have both translational and rotational motion. To determine the acceleration of the pulley, we can use the equation for rotational motion, τ=Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The torque acting on the pulley is equal to the tension in the string (T1) multiplied by the radius of the pulley (R). The moment of inertia of a solid disk is given by 1/2MR^2. So, we can write the equation:

τ = TR = Iα

Since the pulley and the blocks are connected by the string, their accelerations are the same. So, we can equate the linear acceleration (a) and the angular acceleration (α) and solve for a to get:

a = αR

Substituting this value of a in the equations for block m1 and m2, we get:

m1αR = T1 - μm1g
m2αR = m2gsin(θ) - T2 - μm