Calculate Acceleration of Mass m3 with Non-massless Pulley

[nahya]

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.55. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?
---
because of the constraint that the string does not slip on the pulley, the angular acceleration of the pulley is equal to the acceleration at which mass m3 falls, right?
alpha = torque / I
by definition, I in this case is 1/2 * m2 * R^2,
torque = I * alpha = I * a, and
T + µ*m1 - m3g = m3a, right?
but how do i find out what the tension would be?
wouldn't i need the acceleration to find the tension?
but what I'm trying to find out is the acceleration... bleh.
the next two questions ask about the horizontal tension and the vertical tension (of the rope). so I'm guessing the two tensions are different...
now the problem just got more confusing.

am i even heading to the right direction?
thanks.

 

[Andrew Mason]

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.55. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?
---
because of the constraint that the string does not slip on the pulley, the angular acceleration of the pulley is equal to the acceleration at which mass m3 falls, right?
alpha = torque / I
by definition, I in this case is 1/2 * m2 * R^2,
torque = I * alpha = I * a, and
T + µ*m1 - m3g = m3a, right?
but how do i find out what the tension would be?
wouldn't i need the acceleration to find the tension?
but what I'm trying to find out is the acceleration... bleh.
the next two questions ask about the horizontal tension and the vertical tension (of the rope). so I'm guessing the two tensions are different...
now the problem just got more confusing.

am i even heading to the right direction?
thanks.

This net force is supplied by the difference in tension in the rope across the pulley. The tension in the rope between the pulley and m1 supplies the acceleration to m1 and overcomes the force of friction as well. The additional tension in the rope between m3 and the pulley provides the torque to the pulley.

So express the acceleration of each part in terms of the tensions in the rope. Work that out to find a.

AM

 

[kyle8921]

As a scientist, my response would be to approach this problem using Newton's Second Law, which states that the net force on an object is equal to its mass multiplied by its acceleration (ΣF = ma). In this case, the net force acting on mass m3 is the tension in the string (T) minus the weight of the mass (m3g). Therefore, we can set up the equation:

ma = T - m3g

Next, we can look at the forces acting on the pulley. The tension in the string is also the force that causes the pulley to rotate, and the frictional force (µm1) is acting in the opposite direction. Therefore, the net torque on the pulley is:

Στ = Iα = TR - µm1R

Where I is the moment of inertia of the pulley (1/2*m2*R^2) and R is the radius of the pulley. Since we know that the angular acceleration of the pulley (α) is equal to the linear acceleration of mass m3, we can set these two equations equal to each other:

ma = TR - µm1R

Now, we can solve for the acceleration (a):

a = (TR - µm1R)/m3

To find the tension in the string (T), we can substitute this value for acceleration into the first equation:

T = ma + m3g

Finally, to find the horizontal and vertical tensions in the string, we can use basic trigonometry and the fact that the angle between the string and the horizontal is the same as the angle between the string and the vertical (since the string does not slip on the pulley):

Th = Tcosθ = (ma + m3g)cosθ

Tv = Tsinθ = (ma + m3g)sinθ

Overall, the key to solving this problem is to use Newton's Second Law and to consider the forces and torque acting on each object in the system.