What is the Net Force on a Pulley System with Unequal Masses?

[rash219]

Another Pulley Question !

Homework Statement

This question is with a diagram. There are two masses at the end of each rope on a pulley..according to the diagram Mass-2 (right) is greater than Mass-1 (left). Note that gravity acts on both masses. Thats all they give..the questions are
a) Determine the Net Force acting on the system.
b) What is the total mass of the system.
c) Find the acceleration of the system.

b]2. Homework Equations [/b]

Weight = Mass * Gravity
Force = Mass * Acceleration

The Attempt at a Solution

b)

W1 = m1*g
m1 = W1/g ------(1)

W2 = m2*g
m2 = W2/g ------(2)

(1) + (2) = (W1 + W2)/g

a)

\SigmaF1 = m1*a
T - W1 = (W1/g)*a
T = (W1/g)*a + W1 -------(3)

\SigmaF2 = m2*a
W2 - T = (W2/g)*a
-T = (W2/g)*a - W2
T = W2 - (W2/g)*a -------(4)

Ok now i am all confused with all these equations how do i relate it question (a) and (c) ?

 

[Doc Al]

b)

W1 = m1*g
m1 = W1/g ------(1)

W2 = m2*g
m2 = W2/g ------(2)

(1) + (2) = (W1 + W2)/g

Why don't you express things in terms of mass (m_1 + m_2) instead of weight?

a)

\SigmaF1 = m1*a
T - W1 = (W1/g)*a
T = (W1/g)*a + W1 -------(3)

\SigmaF2 = m2*a
W2 - T = (W2/g)*a
-T = (W2/g)*a - W2
T = W2 - (W2/g)*a -------(4)

I assume this is (c). Again, why use weight? In any case, you should be able to solve for the two unknowns: T & a. (Your equations are correct.)

As far as answering part (a), what forces act on the system? (I assume the system is the two masses.) Add them up!

 

[Mr Virtual]

As far as answering part (a), what forces act on the system? (I assume the system is the two masses.) Add them up!

Actually it is the Net force acting on the system due to g that we have to calculate, which (as far as I know) means we will have to subtract the lighter weight from the heavier weight. This net force will act downwards.
I am, though, a bit doubtful about the acc of system. The lighter mass is accelerating up and heavier one down. I think I am missing something here.

Mr V

 

[Doc Al]

Actually it is the Net force acting on the system due to g that we have to calculate, which (as far as I know) means we will have to subtract the lighter weight from the heavier weight. This net force will act downwards.

You need to calculate the net force on the masses due to all forces, not just due to gravity. Gravity is not the only force acting on the masses. (This does turn out to be the difference between the weights.)

I am, though, a bit doubtful about the acc of system. The lighter mass is accelerating up and heavier one down. I think I am missing something here.

The masses are attached via the rope; the magnitude of their acceleration is the same.

 

[rash219]

Ok i revised all my calculation and came up with this attempt i assuming that (b) is right from my attempt i would calculate (c) as below

http://img151.imagevenue.com/loc682/th_98713_dig_122_682lo.jpg

For mass 1

\SigmaF = m_1 * a
T - W1 = m_1*a
T - (m_1*g) = (m_1*a) ------(3)

For mass 2

\SigmaF = m_2 * a
W2 - T = m_2*a
(m_2*g) - T = (m_2*a) ------(3)

(3) + (4) => (m_2*g) - (m_1*g) = (m_2 + m_1) * a

therefore a = [(m_2 - m_1) / (m_2 + m_1)] * g ---------------(5)

Similarly (a)

\SigmaF = m * a
\SigmaF = (m_1 + m_2) * a
From (5)
\SigmaF = (m_2 - m_1)*g

Would this be correct ??

 

[Doc Al]

For mass 1

\SigmaF = m_1 * a
T - W1 = m_1*a
T - (m_1*g) = (m_1*a) ------(3)

For mass 2

\SigmaF = m_2 * a
W2 - T = m_2*a
(m_2*g) - T = (m_2*a) ------(3)

(3) + (4) => (m_2*g) - (m_1*g) = (m_2 + m_1) * a

therefore a = [(m_2 - m_1) / (m_2 + m_1)] * g ---------------(5)

Perfectly correct.

Similarly (a)

\SigmaF = m * a
\SigmaF = (m_1 + m_2) * a
From (5)
\SigmaF = (m_2 - m_1)*g

That answer is correct, but there are some subtleties.

You can look at m_1 and m_2 together by considering the motion to be along a line parallel to the rope. In that case, your equation F = (m_1 + m_2) * a would apply. Be careful with this method. (It's fine if you are aware of what you are doing. The way the problem is worded sounds like they expect this method.)

Looked at in a more typical way, realize that m_1 and m_2 have opposite accelerations. So F = (m_1 + m_2) * a does not apply (at least where "a" is the magnitude of the acceleration of each mass). Instead, find the net force directly by considering all forces on the masses. (Include the tension, which you'll have to solve for.) Of course, this is just a different way of getting the same answer.

 

[rash219]

so in other words are you trying to say (a) Determine the Net Force acting on the system...is the same thing as finding the tension...?

 

[Doc Al]

No. I'm saying that gravity is not the only force acting on the two masses--the rope also pulls on them.