# Pulley Problem: Tension on one side, Weight on another

by frostedpoptar
Tags: pulley, tension, weight
 P: 15 1. The problem statement, all variables and given/known data problem #62 2. Relevant equations Torque = F * radius Tangential accel = radius * alpha w = mg Kinematic equations for linear motion 3. The attempt at a solution I tried to find the net torque and go from there. For clockwise torque I got T = 10 * .25 = 2.5 Nm For ccw torque I got T = 1.5g * .25 = 3.675 Nm For net torque I then get ccwT - cwT = 1.175 Nm I used that with Torque = I * alpha to find alpha. 1.175 = .5 * m * r^2 * alpha 1.175 = .0625 * alpha 18.8 = alpha Then to get tangential accel I did Accel = .25 * 18.8 accel = 4.7 m/s^2 Plug that into kinematics .3 = .5 * 4.7 * t^2 .357 s = t This is wrong however, the answer is .56 s. Can anyone tell me where I went wrong? Thanks.
 P: 15 I also tried to do F = ma and in turn, 4.7 = 1.5a I then substituted that a into kinematics and got something like .43 s. Still wrong.
HW Helper
P: 2,318
 Quote by frostedpoptar I also tried to do F = ma and in turn, 4.7 = 1.5a I then substituted that a into kinematics and got something like .43 s. Still wrong.
You are also accelerating the pully - we commonly use mass-less, frictionless pulleys in this type of problem - so it will take longer.

 P: 15 Pulley Problem: Tension on one side, Weight on another Then how do I go about doing this?
 P: 18 The counterclockwise tension equals T*r, where T=tension in the string and r is the radius of the pulley. Because the mass is accelerating downward, you cannot say that tension on the string on the "left side" of the pulley is equal to the mass. If that were true, the mass wouldn't accelerate. This is where you went wrong. Instead, you have to write a system of equations and solve for the acceleration and, subsequently, time. The equations you write should be for the net torque (in terms of tensions) on the pulley and the net force on the block. The tension on the right side is 10N, so you only have two unknowns, A and T1 (If T1 is the tension of the left string) and two equations. Remember that the linear acceleration of the pulley equals the linear acceleration of the weight. After you use the dynamics equations for acceleration, use the kinematics. I got 0.56 seconds for my answer this way.
 P: 15 Houdini, thanks so much for the reply. I see what you're getting at. It's probably just me but I can't seem to apply it. Torque cw = T * .25m Torque ccw = mg - T * .25m? I'm probably making the same mistake you mentioned in your first paragraph though. Can you please elaborate? It's greatly appreciated. Physics Final tomorrow :)
P: 18
 Quote by frostedpoptar Houdini, thanks so much for the reply. I see what you're getting at. It's probably just me but I can't seem to apply it. Torque cw = T * .25m Torque ccw = mg - T * .25m? I'm probably making the same mistake you mentioned in your first paragraph though. Can you please elaborate? It's greatly appreciated. Physics Final tomorrow :)
Okay, well, remember to just take it one step at a time. It looks like you've make the force and the torque the same thing, but you should leave them apart and find one equation for torque and one for force.

Torquenet=T1*r-T2*r=T1*r-10*r

So all I've done here is taken the two torques and subtracted them. You don't know T1 yet, but you do know it's relationship to the net torque, and the net torque's relationship to the acceleration. And T2 is just the 10 Newtons

T1*r-10*r=Iaangular=Ia/r

Divide by r

T1-10=Ia/r^2=(1/2Mr^2*a)/r^2=(ma/2)

So now you have a pretty easy equation for A and T1 (T1-10=ma/2; this is the "torques equation", but after our manipulation it's more of a "tensions equation" because T1 is just the tension on the string. I'll still refer to it as the "torques equation". Remember that the torque=tension*radius, or t=T*r)

Now you need the pretty simple equation for the forces

Forcenet=-mblocka=T1-mblockg

This follows from the tension pulling up on the block (Which you have never known), and the block's weight pulling down. The result of adding them is the sought after acceleration of the block times it's mass, -mblocka.

You'll notice that A in the force equation and the A in the torques equation should be the same, now just solve for T1 (Easier from the torque equation), substitute it into the other equation, and go to town. And no problem, I hope that helps.

You should get a=(mblock*g-10)/(mblock+I/r^2),
 P: 15 Just one thing I'm confused about in your explanation now In "T1-10=Ia/r^2=(1/2Mr^2*a)/r^2=(ma/2)" how did you get ma/2? When I simplify this I get T1-10 = ((1/2)(M)(r^2)(a))/(r^2) T1 - 10 = ((1/2)(2)(r^2)(a))/(r^2) 1/2 cancels the 2 T1 - 10 = ((r^2)(a))/(r^2) r^2's cancel T1-10 = a Where did I go wrong?
 P: 15 Oh nevermind I see what you did. You just didnt substitute the 2 for m.