Pulley, two objects, frictionless plane; find acceleration and cable tension

[emr13]

Homework Statement

A box of mass m₁ is hanging from a cable of negligible mass, The cable rounds a pulley, where it is connected to a second box of mass m₂ on a frictionless plane. The angle of the inclined plane is \theta as measured from the flat ground. Both boxes are initially at rest. Find the acceleration of the boxes and the tension in the cable.

Homework Equations

\overline{F}=m\overline{a}

\overline{n}+\overline{F}_g=m\overline{a}

The Attempt at a Solution

Object A had mass m₁ and object B has mass m₂.

I know that gravity and the cable tension are both affecting object A, and the cable tension, gravity, and a normal force from the plane are affecting object B. Therefore, I have:

Object A: T_A + F_gA = m₁a
Object B: T_B +F_gB + n = m₂a

(where T, F, and a are all vectors)

I tried drawing free-body diagrams, and I just get lost drawing all of the equations...I know I need to make component vectors, but there is a lot to the problem, and I'm feeling a little overwhelmed.

Is this stuff correct so far? Thanks.

 

[Doc Al]

What forces act on A? What forces act on B? (For B, all you care about are forces parallel to the incline.)

 

[emr13]

Gravity and the cable tension are the forces acting on A.

Gravity, the cable tension, and the normal force (the plane) act on force B.

 

[Doc Al]

Gravity and the cable tension are the forces acting on A.

Good. Set up an equation with Newton's 2nd law.

Gravity, the cable tension, and the normal force (the plane) act on force B.

What are their components parallel to the incline? Set up an equation with Newton's 2nd law.

You'll get two equations with two unknowns. Careful with signs. (If one block goes up, the other must go down.)

 

[emr13]

Assuming A is heavier than B...

for A, I get m₁g - T = m₁a

The parallel forces for object B are F_gx and the T force... so I guess for B I get
T-m₂gsin\theta = m₂a.

Am I ignoring the normal force because the object doesn't move in that direction so it doesn't affect the acceleration?

If the above equations are right, then I can get the following equation for acceleration by making T= m₂a + m₂gsin(theta), and substituting it in for the T in object A's equation:

a=m₁g-m₂gsin(theta) over m₁+m₂

(sorry, formatting wouldn't work right for some reason)

 

[Doc Al]

Assuming A is heavier than B...

for A, I get m₁g - T = m₁a

The parallel forces for object B are F_gx and the T force... so I guess for B I get
T-m₂gsin\theta = m₂a.

Excellent.

Am I ignoring the normal force because the object doesn't move in that direction so it doesn't affect the acceleration?

Right. You can consider the normal force if you like, but since the acceleration in that direction is zero, you'll just end up with N = m₂gcosθ. You don't need it.

If the above equations are right, then I can get the following equation for acceleration by making T= m₂a + m₂gsin\theta, and substituting it in for the T in object A's equation:

a=m₁g-m₂gsin\theta
m₁+m₂

There are several ways to combine the two equations; that way is perfectly fine. I assume you meant to write:
a = (m₁g - m₂gsinθ)/(m₁ + m₂)

Another way is simply to add the two equations. That way the T from one cancels the -T from the other.

 

[emr13]

Yes, that it what I meant for the acceleration.

To find the cable tension, I substituted acceleration back into one of the equations and got:

T=m₁g - m₁²g+m₂gsinθ/m₁+m₂

that looks really gross...is it still correct?

And thank you so much for all your help.

 

[Doc Al]

Yes, that it what I meant for the acceleration.

To find the cable tension, I substituted acceleration back into one of the equations and got:

T=m₁g - m₁²g+m₂gsinθ/m₁+m₂

that looks really gross...is it still correct?

It looks a bit off to me, but it may be formatting again. In any case, your approach is correct. If you do it over more carefully, you can simplify it a bit so it doesn't look quite so ugly. (Hint: Express everything over m₁ + m₂.)

 

[emr13]

What about this?

T=(m₁m₂g(1+sinθ)) / (m₁+m₂)

 

[Doc Al]

What about this?

T=(m₁m₂g(1+sinθ)) / (m₁+m₂)

Looks good!

 

[emr13]

Awesome. Thank you so, so much.