What is the Solution to an Inclined Plane with Pulley Problem?

[wy125]

Homework Statement

The included image shows the problem in its entirety. The problem is from a physics review book that I have which also includes the answer, but the explanation is beyond me. I was hoping someone could walk me through the solution.

T, the rope tension (constant throughout the rope)
m_b - the mass of the block = 20 kg
m_p - the mass of the lower pulley = unknown
F as the net force acting on the pulley (shown in image) = 100 N
$$\vartheta$$, the angle of incline = 30 degrees


http://img705.imageshack.us/img705/7307/pullyramp.png

Homework Equations

The Attempt at a Solution

Here's how I tackled this:

(1) a_b = 2 a_p The acceleration of the block is twice that of the pulley, since moving the pulley a distance d results in the block moving a distance 2d

A free-body diagram for the block shows that the forces pushing the block up or down the ramp are T, the rope tension pushing up, and gravity acting to move the block down the ramp.

So, taking the direction up the block to be positive we have:

(2) T - mgsin$$\theta$$ = m_ba_b

Similarly for the bottom pulley we have (taking down to be positive)

(3) 2T - F = m_pa_p


That's as far as I get. I have 4 unknowns: T, m_p, a_p, a_b

but only 3 equations.

Any insight or help will be appreciated.

 

[magwas]

Either they meant the pulley massless, or F being its gravitational force.
I believe in other cases the problem only solvable as a function of mass of the pulley.
So let's rewrite (2) as
$$T = m_{b} g sin(\theta)$$

I know you can solve it from here.

 

[wy125]

So let's rewrite (2) as
$$T = m_{b} g sin(\theta)$$

If we let that be the case, then the tension in the rope is balancing out the gravitational force along the ramp, which means that there is no acceleration. From the solution given, I can tell you that there is a nonzero acceleration. Have I missed something in your analysis?

Thank you for your quick reply by the way!

 

[magwas]

Either g <> 10 m/s^2, but 9.81 m/s^2, or we should figure out where the other end of the rope fixed, or something I cannot figure out.

 

[magwas]

ahh.
T=F/2=50N //it is (3) with massless pulley
Fgrav=m*g*sin(theta)=~100N (using g=10) // this is not the rope tension, just the gravitational force in the direction of rope

Facceleration=T-Fgrav =~ 50N

 

[wy125]

I'm not sure I'm following you exactly. What would you say the acceleration of the block is? 50N/20kg = 2.5 m/s? If so, that's not the stated answer. Of course the stated answer could be wrong. That is the answer i got btw the first time I tried this, but I don't think it's right to assume that T = 1/2 F because the system is accelerating. I'm not sure though -- that's why I'm here

 

[magwas]

Yes, I would say 2.5 m/s.
And I think that T=1/2F whether the block accelerates or not.
I can be wrong, of course.
What is the stated answer?

 

[wy125]

The stated answer is 5 m/s up the incline. By the way the acceleration we get from using T= 50 N results in acceleration of the block down the plane.