How Do You Calculate Tension in a Two-Block Pulley System?

[Bones]

Homework Statement

Two blocks are connected by a light string passing over a pulley of radius 0.35 m and moment of inertia I. The blocks move (towards the right) with an acceleration of 1.50 m/s2 along their frictionless inclines (see the figure).

http://www.webassign.net/gianpse4/10-54alt.gif

Determine FTA and FTB, the tensions in the two parts of the string.

Find the net torque (magnitude only) acting on the pulley, and determine its moment of inertia, I.

Homework Equations

The Attempt at a Solution

Is there a separate tension for each piece of string and what are the equations if that is the case. If it is not the case, then what is the equation to find the tension?? Please Help!

 

[Doc Al]

Is there a separate tension for each piece of string and what are the equations if that is the case.

The tension must be different on each side of the pulley, otherwise there'd be no torque to turn it.

The equations you need are Newton's 2nd law. Apply it (separately) to each mass and the pulley, then combine those three equations to solve for your unknowns.

 

[Bones]

I am not sure how to do that with the tension and all.

 

[mathmate]

Indeed the tensions are different!
There are actually three systems here, namely
1. Block A which accelerates up the incline with an acceleration of 1.5 m/sec².
2. The pulley which has a tangential acceleration of 1.5 m/sec² at r=0,35 and moment of inertia I
3. Block B which descends with an acceleration of 1.5 m/sec² down the incline.

Systems 1 and 3 are related by Newton's law of motion
F=ma.
System 2 is related by the analogous equation
\tau=I\alpha
where \tau=torque and
\alpha=angular acceleration=a/r
r=radius of pulley,
a=tangential acceleration at the rim of the pulley

So you have three equations and unknowns, Ta, Tb and I, and you should have no problems solving for them.

 

[Doc Al]

I am not sure how to do that with the tension and all.

Start by picking one of the systems and identifying the forces acting on it. Drawing a diagram would be wise. Then apply Newton's 2nd law: ∑F = ma (or ∑τ = Iα, for the pulley).

 

[Bones]

Block A: F=8.0kg*1.50m/s^2
Block B: F=10.0kg*1.50m/s^2
Pulley: T=(2/5Mr^2)(1.50m/s^2)

Is that correct??

 

[phynoob]

can we tilt the diagram,61 degree clock wise such that the gravitational force that is acting on block b is 9.8m/s^2. and there will be a downward acceleration for block a?
can we use that approach to solve this question as well?

 

[Bones]

I have no idea.

 

[mathmate]

What you have got are the net forces (mass +/- string tension) acting on the mass.
For the pulley, you'd need to convert the linear acceleration into angular acceleration using the radius. I is one of the three unknowns, since you do not know the mass of the pulley.

The downward force of the 8 and 10 kg masses can be resolved into two components, one parallel to the inclined plane, and the other, normal to it.

The normal force does not do anything, as the inclines are frictionless. The forces that are parallel to the inclined planes are the ones causing the acceleration. Do not forget to add/subtract the tension of the string from the force parallel to the incline.

 

[Bones]

Net force=ma=force of tension-mg

Block A: (8kg)(1.50m/s^2)=Ft-(8kg)(9.8m/s^2)
Block B: (10kg)(1.59m/s^2)=Ft-(10kg)(9.8m/s^2)

Is that what you mean?

 

[Bones]

Correction:

Block A: Ft-(8kg)(9.8m/s^2)sin32=(8kg)(1.50m/s^2) Ft=53.55N

Block B: -Ft+(10kg)(9.8m/s^2)sin61=(10kg)(1.5m/s^2) Ft=70.71N

 

[Bones]

So to find the net torque: Block A: (0.35m)(53.55N)=18.74 Block B: -(0.35m)(70.71N)=-24.75
Net Torque: 18.74-24.75=-6.01N*m

 

[Bones]

Moment of inertia: t=I(alpha) or t=I(a/r)
-6.01N*m=I(1.50m/s^2)/(0.35m)
I=-11.45kg*m^2

What am I doing wrong here?

 

[Doc Al]

Moment of inertia: t=I(alpha) or t=I(a/r)
-6.01N*m=I(1.50m/s^2)/(0.35m)

This is fine, except for the minus sign. You just need the magnitude of the net torque.

I=-11.45kg*m^2

Redo your calculation; you made an arithmetic error.

 

[Bones]

Ok, I see what I did.

Thanks for the help ;)