1. ## Pulley question

Hi. I've started this problem, and have managed as far as proving the result for acceleration. If someone could finish it, or explain how to finish it so that I can finish it myself, that would be great. Thanks.

A planed is inclined at an angle arctan ¾ to the horizontal and a small, smooth, light pulley P is fixed to the top of the place. A string, APB, passes over the pulley. A particle of mass m1 is attached to the string at A and tests on the inclined place with AP parallel to a line of greatest slope in the place. A particle of mass m2, where m2 > m1, is attached to the string at B and hangs freely with BP vertical. The coefficient of friction between the particle at A and the plane is ½.

The system is released from rest with the string taut. Show that the acceleration of the particles is ((m2 – m1)g)/(m2 + m1).

At a time T after release, the string breaks. Given that the particle at A does not reach the pulley at any point in its motion, find an expression in terms of T for the time after release at which the particle at A reaches its maximum height. It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at A to descend from its point of maximum height to the point at which it was released. Find the ratio m1 : m2.

2. Originally Posted by Thomas154321
Hi. I've started this problem, and have managed as far as proving the result for acceleration. If someone could finish it, or explain how to finish it so that I can finish it myself, that would be great. Thanks.

A planed is inclined at an angle arctan ¾ to the horizontal and a small, smooth, light pulley P is fixed to the top of the place. A string, APB, passes over the pulley. A particle of mass m1 is attached to the string at A and tests on the inclined place with AP parallel to a line of greatest slope in the place. A particle of mass m2, where m2 > m1, is attached to the string at B and hangs freely with BP vertical. The coefficient of friction between the particle at A and the plane is ½.

The system is released from rest with the string taut. Show that the acceleration of the particles is ((m2 – m1)g)/(m2 + m1).
The solution you have given for the first part is not correct.

Who gives an angle in terms of arctan in Physics? Let's get that out of the way now and say that the angle of incline is theta = 36.87 degrees or so. (I would use a much longer decimal expression for calculations.)

We need to do a Free Body Diagram for each mass. So.

Mass 1: There is a tension T acting on m1 directed up the plane, a normal force N acting perpendicular to the plane, a weight w1 acting directed straight downward, and a friction force f acting down the plane. (We can tell m1 will be moving up the plane since m2 > m1.) My coordinate system will be +x up the plane and +y in the direction of the normal force. (I usually pick one of the directions to be in the direction of the acceleration if there is one. I'll explain why in a minute.)

Now do Newton's 2nd Law in each coordinate direction:
SumFx = T - f - w1*sin(theta) = m1*a
SumFy = N - w1*cos(theta) = 0 (since m1 does not accelerate in this direction.)

Of course we have that w1 = m1*g and f = (mu)*N so these equations become:
T - (mu)N - m1*g*sin(theta) = m1*a
N - m1*g*cos(theta) = 0

We are assuming we know m1 and m2, we know (mu), g, and theta. So this is 2 equations in 3 unknowns. We need another equation to solve the system. So let's move on to m2.

Mass 2: We have a tension T acting upward (equal to the T in the FBD for m1 because we are using an ideal string) and a weight w2 acting downward. I am going to choose a direction for +y to be downward. Why? Because we chose a positive direction to be up the plane in the m1 FBD and we need to keep the "sign" of the direction of acceleration consistent. If m1 accelerates up and is positive, then m2 accelerates down and should also be positive. (We can choose positive upward here, but that will induce a negative sign on the acceleration in Newton's 2nd Law. My choice avoids this problem.)

Now do Newton's 2nd Law:
SumFy = w2 - T = m2*a (Note the signs on w2 and T! Positive is downward here!)
Again w2 = m2*g so we get

m2*g - T = m2*a

So we have a system of three equations in three unknowns (a, T, and N):
T - (mu)N - m1*g*sin(theta) = m1*a
N - m1*g*cos(theta) = 0
m2*g - T = m2*a

We can immediately solve the middle equation for N and substitute that in the top equation:
N = m1*g*cos(theta)

T - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a
m2*g - T = m2*a

We want an expression for a, so I am going to solve the bottom equation for T and substitute that into the top equation:
T = m2*g - m2*a

m2*g - m2*a - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a

m2*g - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a + m2*a

We can factor the common "a" on the RHS and then divide both sides of the equation by m1 + m2:

a = [m2*g - m1*g*{ (mu)*cos(theta) + sin(theta) }]/(m1 + m2)

-Dan

3. Originally Posted by Thomas154321
At a time T after release, the string breaks. Given that the particle at A does not reach the pulley at any point in its motion, find an expression in terms of T for the time after release at which the particle at A reaches its maximum height.
We know from the previous post what the acceleration is. We also know that (mu) = 0.5 and the angle theta = 36.87 degrees (or so).

The acceleration is constant so long as the string is attached to m1. We know that m1 starts from rest. So if the string breaks at a time T we know that the speed of m1 at time T will be v0 = aT. (I'm using v0 in deference to the next step.)

At time T the string breaks and thus the acceleration changes. Removing the tension force there are 2 remaining forces in the m1 FBD directed along the slope: w1*sin(theta) and f = (mu)N, both of these acting down the incline. So the net force on m1 after time T is now:
SumFx = -m1*g*sin(theta) - (mu)*m1*g*cos(theta) = m1*A (Calling the new acceleration A.)

Thus
A = -g[sin(theta) + (mu)*cos(theta)]
v0 = aT (with "a" from the previous post)

So reset the clock to be t = 0 when the string breaks. The equation of motion of m1 will be:
x = v0*t + (1/2)A*t^2

The maximum height is obtained when dx/dt = 0, or when 0 = v0 + At, or:
t = -v0/A

t = aT/{g[sin(theta) + (mu)*cos(theta)]}

t = {[m2*g - m1*g*{ (mu)*cos(theta) + sin(theta) }]T}/(m1 + m2)}/{g[sin(theta) + (mu)*cos(theta)]}

t = {[m2*g - m1*g*{ (mu)*cos(theta) + sin(theta) }]T/{(m1 + m2)*g[sin(theta) + (mu)*cos(theta)]}

(Which is NOT going to simplify into anything nice.)

We are going to need the distance m1 slides in this time to do the next part. This will be x, and I'm going to solve for it using:
v^2 = v0^2 + 2A(x - x0)

We know that v = 0 (m1 is at max. height) and x0 = 0 here by definition. Thus
x = -v0^2/(2A)

x = (aT)^2/(2*g[sin(theta) + (mu)*cos(theta)])

(You can sub in the "a" value. It doesn't simplify either.)

-Dan

4. Originally Posted by Thomas154321
It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at A to descend from its point of maximum height to the point at which it was released. Find the ratio m1 : m2.
There's a problem with this one too.

Take a look at the m1 FBD, without the tension force. This was done for an object sliding up the incline so the friction force was directed downward. Now m1 is going to slide down the incline so the friction force will be directed up the incline. So the new sum of forces on m1 will be:
SumFx = -m1*g*sin(theta) + (mu)*m1*g*cos(theta) = m1*B (I've run out of a's! So I'm calling this third acceleration "B.")

Thus
B = g[(mu)*cos(theta) - sin(theta)]

m1 starts from rest and slides over a distance x. We need to calculate the time to do this.

Here we have v0 = 0, so x = (1/2)Bt^2. Thus

t = sqrt(2x/B)

This will be the same as t from the previous post and we are supposed to use this information to solve for the ratio m1/m2.

Define D = [(mu)*cos(theta) - sin(theta)]

We have that
a = [m2*g - m1*g*D]/(m1 + m2)

t = [m2*g - m1*g*D]T/{(m1 + m2)*g*D}

x = (aT)^2/(2*g*D) = T^2*[m2*g - m1*g*D]^2/{2*g*D*(m1 + m2)^2}

and the new t we got in this post is:

t = sqrt(2x/B) = sqrt(2*T^2*[m2*g - m1*g*D]^2/{2*g*D*g[(mu)*cos(theta) - sin(theta)]*(m1 + m2)^2}

If we set E = [(mu)*cos(theta) - sin(theta)] the new t becomes:
t = sqrt(2*T^2*[m2*g - m1*g*D]^2/{2*g*D*g*E*(m1 + m2)^2}
= {T*[m2 - m1*D}/(m1 + m2)]\sqrt{DE}

Now, t = t, so
[m2*g - m1*g*D]T/{(m1 + m2)*g*D} = {T*[m2 - m1*D]}/(m1 + m2)]/sqrt{DE}

Cancelling some common terms I get:
1 = 1/sqrt{DE}.

I can't find a mistake (I may have made one. Please check.) so this is a condition on (mu) and theta. Whether this condition is realistic or not (I think the time condition given for this problem is not realistic, so I'm betting we have a contradiction here) the point is that the masses have cancelled out completely!

My conclusion is that I don't think we can get the ratio.

-Dan

5. Originally Posted by Thomas154321
The system is released from rest with the string taut. Show that the acceleration of the particles is ((m2 – m1)g)/(m2 + m1).
I happen to recognize this acceleration: it's for the Atwood machine, which is simply two masses suspended on either side of a pulley. How much do you want to bet that you can do this whole problem on an Atwood's machine? (I'm guessing someone tried to construct a new problem similar to one they found in a book and didn't realize how nasty it was going to get with the inclined plane replacing the Atwood's machine.)

-Dan