Calculating tensions and acceleration

[shepherd882]

Homework Statement

A mass M1 is sliding across a table with coefficient of kinetic friction μk. A string is tied to this mass and runs over a pulley, drops vertically and is tied to another mass M2 which is falling. The pulley is connected to the table by a support. The pulley is a solid cylinder of mass M3, radius R. See figure. As the figure indicates, you may assume that the string is tied to the midpoint of both masses, that it runs horizontally from M1 to the pulley and vertically from the pulley to M2. You may also assume that the mass M1 is not tipping over as it slides.

a) Assume the string slips over the pulley with zero friction, and that the pulley is not rotating. Calculate the magnitude of the acceleration of M1 and M2. (This is the same for both masses.) Also find the tensions T1 in the string attached to M1 and T2 in the string attached to M2. Express your answers in terms of g and the various parameters given in the question.
b) Now assume that the string runs over the pulley without slipping, making the pulley rotate with increasing angular speed. Assume the pulley is attached to the table by a support which allows it to rotate on an axle with no friction between the pulley and axle. Calculate the magnitude of the acceleration of the masses M1 and M2 in this case. Also calculate the tensions T1 and T2 in this case.

Thanks for you help!

Homework Equations

Fnet = ma
Torque net = I*a
I = (MR^2)/2
Ff = uFn

The Attempt at a Solution

a) [/B]Fnet = ma
Fg2-T2+T1-Ff1 = (M1+M2)a
Fg2 - Ff1 = (M1+M2)a
M2g-uM1g = (M1+M2)a
(M2g-uM1g)/(M1+M2) = a

Fnet1 = T1-Ff1 = M1a
T1 = M1a + Ff1
T1 = M1a + uM1g
Fnet2 = Fg2 - T2 = M2a
T2 = Fg2 - M2a
T2 = M2g - M2a

b) Torque = I*a
T2R-T1R = Ia
a = [R^2/(M3R^2)]*(T2-T1)
a = (T2-T1)(2/M3)

T1 and T2 same as part a)?

 

[BvU]

T1 and T2 same as part a)?

Simple answer: no. Some difference is needed for ... (fill it in)

[edit] by the way, part a) isn't finished yet !

 

[Chestermiller]

In the first part of your analysis for part (a), the tensions should not be included in the overall tangential force balance because they are internal to the system.

 

[shepherd882]

Simple answer: no. Some difference is needed for ... (fill it in)

[edit] by the way, part a) isn't finished yet !

for part a), would i just substitute acceleration into each of the tension equations then simplify?
also for part b), the difference would be that mass 3 has the normal force at an angle which would add to the fnetx component. am i on the right track?

 

[Chestermiller]

for part a), would i just substitute acceleration into each of the tension equations then simplify?

Yes. You should get the same answer for both tensions. Do you?

also for part b), the difference would be that mass 3 has the normal force at an angle which would add to the fnetx component. am i on the right track?

No. This is not the right approach. You should start by doing a moment balance on the pulley. In part (b), you should have 3 separate balance equations: one for each mass, and one for the pulley.

 

[shepherd882]

Yes. You should get the same answer for both tensions. Do you?
No. This is not the right approach. You should start by doing a moment balance on the pulley. In part (b), you should have 3 separate balance equations: one for each mass, and one for the pulley.

That's what I got for T1 and T2 for part a) ...

 

[shepherd882]

Yes. You should get the same answer for both tensions. Do you?
No. This is not the right approach. You should start by doing a moment balance on the pulley. In part (b), you should have 3 separate balance equations: one for each mass, and one for the pulley.

Is this correct?

 

[Chestermiller]

View attachment 233904
That's what I got for T1 and T2 for part a) ...

That's not what I get. For part (a), I get $T_1=T_2=T$ with $$T=\frac{m_1m_2(1+\mu_k)}{(m_1+m_2)}g$$
This is equivalent to your result for $T_2$ but not for $T_1$

 

[Chestermiller]

View attachment 233905
Is this correct?

I can't read your handwriting, but the equations should read:
$$T_1-m_1g\mu_k=m_1a$$
$$m_2g-T_2=m_2a$$and, for the moment balance on the pulley:
$$T_2R-T_1R=I\frac{a}{R}$$
This gives you 3 equations in the three unknowns $T_1$, $T_2$, and a

 

[shepherd882]

I can't read your handwriting, but the equations should read:
$$T_1-m_1g\mu_k=m_1a$$
$$m_2g-T_2=m_2a$$and, for the moment balance on the pulley:
$$T_2R-T_1R=I\frac{a}{R}$$
This gives you 3 equations in the three unknowns $T_1$, $T_2$, and a

I got those equations

How do you simplify this...?

 

[Chestermiller]

I got those equations
View attachment 233909
How do you simplify this...?

You have one linear algebraic equation in one unknown ("a"); are you really saying that you don't know how to solve this equation for the unknown?