Mechanics Pulley question

[BigCheese234]

Homework Statement

A mass 4m lying on a rough horizontal table is connected by a string,
passing over a pulley, to another mass 3 m hanging freely. The pulley (regarded
as a disk) has mass 4 m and rotates freely about a polar axis parallel to the
edge of the table. The coefficient of kinetic friction between the mass and the
table is μ=3/8. Assuming that the string does not slip on the pulley, Find the
common acceleration of the masses and show that the ratio of the tensions in
the string on either side of the pulley is T1 : T2 = 13 : 15. (You may assume
that the moment of inertia of a disk of radius r and mass m about a polar axis
through the disk is I =1/2 M r^2)

[D] Attempt
acceleration

4mg-t=4a
t-3mg=3A

1g=7a
10=7a
a=10/7
a=1.428
Tension 2:
4m(g)=4m(10)=40m

3/8=T2/40 , 120/8 = T2 , T2=15m

Tension 1:
3(10)=30
am i on the right track?
Thanks for the help have a big exam and this is a common question , having problems with it because the pulley is moving and don't know how it effects the acceleration .solution would be fantastic !:shy:

 

[Villyer]

The force of gravity on the hanging mass will not only accelerate the system of masses, but also cause the pulley to rotate.

 

[BigCheese234]

Thanks for your help ! so the pulley does not effect the common acceleration? does the pulley due to its mass have any bearing on the tension of the hanging mass?

 

[Villyer]

It does effect the acceleration of the mass system because as the mass drops, energy is transferred into the pulley as well, instead of remaining in the system.

 

[Nickie]

Yes, you are on the right track! To solve this problem, you can use Newton's Second Law, which states that the net force on an object is equal to its mass times its acceleration. In this case, you have two masses connected by a string, and the pulley is also involved.

First, let's consider the mass 4m on the table. The forces acting on it are its weight (4mg) and the tension in the string (T1). The friction force between the mass and the table can be calculated as μN, where μ is the coefficient of kinetic friction and N is the normal force (in this case, equal to the weight of the mass). So, we have the following equation for the mass 4m:

4mg - T1 - (μN) = 4ma

Next, let's consider the mass 3m hanging freely. The forces acting on it are its weight (3mg) and the tension in the string (T2). So, we have the following equation for the mass 3m:

T2 - 3mg = 3ma

Since the string is not slipping on the pulley, the acceleration of both masses must be the same (a). We can use this fact to eliminate a from the two equations above. We can also use the moment of inertia of a disk (I = 1/2 Mr^2) to relate the acceleration of the pulley to the tension in the string. The pulley is rotating, so we can use the equation for rotational motion:

T1 - T2 = Iα

where α is the angular acceleration of the pulley, related to the linear acceleration a by α = a/r. Substituting this into the equation above and solving for a, we get:

a = (T1 - T2)/I

Substituting this into the two equations for the masses, we get:

4mg - T1 - (μN) = 4(T1 - T2)/I

T2 - 3mg = 3(T1 - T2)/I

We can simplify these equations to get:

(4 + 3μ)T1 - 4T2 = (4 + 3μ)mg

-3T1 + (3 + 3μ)T2 = -3mg

Now, we can solve these equations for T1 and T2. First,