# Statics - slopes, pulleys and weights

• Apr 12th 2008, 05:52 AM
moolimanj
Statics - slopes, pulleys and weights
Hi Guys, got the following diagram but need to know how I can relate forces in terms of a coordinate system chosen so that I can express the magnitude of each weight in terms of its mas and g.

Note, the system is in equilibrium and m>2Msin(alpha) and the particle A is on the point of slipping up the slope.

Also, how can I use the conditions for equilibrium, together with the propoerties of a model pulley to find the scalar equations relating to the magnitudes of forces?

Finally, does anyone know what the co-efficient of static friction between particle A and the plane looks like?
• Apr 12th 2008, 06:35 AM
moolimanj
Am i right in thinking that the attached is the force diagram for each mass and also a set of suitable co-ordinate axis?
• Apr 12th 2008, 07:28 AM
topsquark
Quote:

Originally Posted by moolimanj
Hi Guys, got the following diagram but need to know how I can relate forces in terms of a coordinate system chosen so that I can express the magnitude of each weight in terms of its mas and g.

Note, the system is in equilibrium and m>2Msin(alpha) and the particle A is on the point of slipping up the slope.

Also, how can I use the conditions for equilibrium, together with the propoerties of a model pulley to find the scalar equations relating to the magnitudes of forces?

Finally, does anyone know what the co-efficient of static friction between particle A and the plane looks like?

Your FBDs look good to me. A good piece of advice is to pick positive directions in the direction that you think the system will accelerate in. (If you are wrong you are wrong, but at least you will be consistent.) If the block on the incline is sliding up the slope, I'd pick a positive direction up the slope. Then, since the block attached to the pulley will be accelerating
downward, I'd choose a positive direction downward. In your case the system is not accelerating so it doesn't matter much, but the directions you chose are consistent anyway.

The model pulley and the ideal string don't affect anything, really. The pully just changes the direction of the tension and the ideal string means the tension in the string is the same along the entire length of the string.

The only error I see in the diagrams is on the diagram for the mass hanging off the pulley: there will be an upward force of 2T since we have since we have two parts of the string pulling upward. (Even though it's the same string.)

I'm not sure what you mean by "what the coefficient of static friction looks like." The coefficient is just a number and doesn't appear on the diagram. If you are referring to the static friction force, I had assumed that is the force F on your diagram, since there is no other force acting in that direction.

As far as applying equilibrium, the mass on the slide is not accelerating, so the sum of the forces in both coordinate directions is 0 N. Similarly the block hanging off the pulley is not accelerating, so the sum of the forces in the upward direction is also 0 N. This will give you a system of three equations to solve for your unknowns.

-Dan
• Apr 12th 2008, 12:45 PM
moolimanj
Thanks Topsquark,

I thought under a model pully, the tensions in the rope are the same. i.e. T1=T2.

Also, if i calculate the components in the coordinate system (assuming the one Ive drawn for mass M has i pointing in the other direction to the one ive drawn, then, will W1=W1(-sin(alpha))i + W1(-sin(alpha))j.

Is this right? I know that for mass m the component in teh coordinate system is W2=-W2j
• Apr 12th 2008, 01:21 PM
topsquark
Quote:

Originally Posted by moolimanj
Thanks Topsquark,

I thought under a model pully, the tensions in the rope are the same. i.e. T1=T2.

Also, if i calculate the components in the coordinate system (assuming the one Ive drawn for mass M has i pointing in the other direction to the one ive drawn, then, will W1=W1(-sin(alpha))i + W1(-sin(alpha))j.

Is this right? I know that for mass m the component in teh coordinate system is W2=-W2j

As to the tensions, yes they are all the same (just perhaps pointing in different directions.)

I'm not sure what you are saying in your other two statements. We apply Newton's 2nd to each diagram individually to set up a system of equations to solve simultaneously.

For example, the x component equation of Newton's 2nd on the mass M (on the slope) gives
$\displaystyle \sum F_x = T - F - Mg~sin(\theta) = 0$
where $\displaystyle \theta$ is the angle of inclination, and F is your maximum static friction force, $\displaystyle \mu _s N$.

-Dan