# What are the magnitude and direction of the friction force?

1. Dec 10, 2011

### skysunsand

1. The problem statement, all variables and given/known data

A 200N block is on a 35 degree incline with a frictionless, massless pulley and a weight on the other side, Fw= 220. The system is in equilibrium.
What are the magnitude and direction of the frictional force on the 200N block?

2. Relevant equations and
3. The attempt at a solution

Ft1 + Ft2 = Fw , so Ft1 and Ft2 are 220N

Fn= 20.4 sin 35 * 9.8 = 112.4

I guess Fw-Fn = Ff, but I don't understand why. Is that always true?
So the answer here would be 108 in regard to the magnitude, though the book wants 105. I guess that's a rounding thing? And it says down the incline, but why?

2. Dec 10, 2011

### Staff: Mentor

What are Ft1 and Ft2? The tensions in the string segments? If so, they do not sum to the weight of the hanging mass. If the pulley is massless and frictionless then they must be the same, yes. If the system is static (in equilibrium) then the tension will be equal to the weight of the hanging mass (as you've written).

Also, if the system is in equilibrium, the net force downslope must equal the net force directed upslope.
What is Fn? It looks like it should be the component of the block's weight that is directed downslope. Is that right? If so, you could have just used the block's weight (200N) and multiplied by the sine of the angle. No need to convert to mass and then back again to weight. As it is you have a numerical error in the result. Check your calculations.
It's true in this case because the system is in equilibrium. Thus the net force acting along the direction of the slope is zero.
Check your math as mentioned above. Friction always opposes the motion or attempted motion. Here the block 'wants' to move upslope, so the friction opposes that attempt and thus directed downslope.

3. Dec 10, 2011

### skysunsand

Yeah, that was my failure to mention the pulley is massless and frictionless. I was never quite sure why they made a point to mention that. Thank you for clarifying that.

Fn is my notation for normal force. I checked my calculations per what you said about converting and I ended up with 114.7, which, when subtracted, does give me 105.3.

How would I know which way the block "wants" to move? Would I just have to look at it and realize that because the weight on the right side is heavier, the block is automatically going to want to move up the slope?

4. Dec 10, 2011

### Staff: Mentor

The normal force doesn't enter into this particular problem because the frictional force is what you will determine from the other forces (so you don't need the normal force to calculate it). What you want is the downlope component of the block's weight. Fortunately, that's what you calculated! (The normal force would be given by mg cos(θ) )
Yes, that or do some preliminary calculations to compare the forces acting on the block.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook