Homework Help: Friction and force at an angle

1. Sep 28, 2010

1. The problem statement, all variables and given/known data
A 27.1-kg block (m1) is on a horizontal surface,
connected to a 5.7-kg block (m2) by a massless string as
shown in the Figure. The pulley is massless and friction-
less. A force of 228.9 N acts on m1 at an angle of 33.5◦.
The coefficient of kinetic friction between m1 and the sur-
face is 0.201. Determine the upward acceleration of m2.

(m2 is dangling off the ledge by the pully connected to m1)

2. Relevant equations

Fg=mg
F=ma
Fk=uk(Fn)

3. The attempt at a solution

I drew all the forces, m1 would have Fg=265.58 going down and Fy=126.338 going up because of the angle it is pulled at. So then I figured Fn would be 265.58-126.338=139.242. And Fx=190.876. And the Fk=.201(139.242)=27.9876. Then m2 would have an Fg=55.86.

I found net force= Fx-Fk-Fgm2=107.028. Then I used F=ma to try and find upward accel of m2. I figured since the force is being applied to m1 and it is connected to m2 I would use the combined mass of both of them and do 107.028=(5.7+27.1)(a) and a came out to be 3.26305m/s^2 but it turns out that was wrong. Can anybody help me find the real answer?

2. Sep 28, 2010

jhae2.718

What forces act on m1 and m2? You have Fg for both blocks, Fn and Fk for m1, but what else?

Think of what connects the blocks; there's a force you are forgetting. (Hint: since the pulley is massless and frictionless, the upward acceleration of m2 is the same as the rightward acceleration of m1.)

3. Sep 28, 2010

Well for me the block is being pushed to the left, friction to the right. But im assuming you meant the upwards acceleration for m1 would be the same as the leftward(for me) acceleration for m1. I knew that already. But I have no idea what force I am missing.

4. Sep 28, 2010

jhae2.718

If the block were being pushed to the left, the hanging block m2 would have a downwards acceleration. From your description, the system sounds like this to me:

Regarding the missing force, are you familiar with tension?

5. Sep 28, 2010

Yea that image looks right, mine was just flipped the other way around but thats fine too. But no I am not sure how to find tension in this problem. Thank you for helping me btw.

6. Sep 28, 2010

jhae2.718

Tension pulls on each end of the rope. When we look at each body in isolation, we "cut" the rope, so there is a tension force acting on each block. By Newton's 3rd law, the forces are equal and opposite, so the tensions on block 1 and block 2 are the same. (This is a bit simplistic of an explanation, and is only true if the pulley is massless and frictionless.)

So, there is a tensile force to the left on block 1 and an upwards tensile force on block 2.

You can solve for tension in block 2 and substitute that into your forces on x for block 1 and then solve for a.

7. Sep 29, 2010

Okay well I tried doing something... I'm not sure if its right.

assuming my force F is 107.028 then,

T-F=m1a

so T=m1a+F

and m2g-T=m2a

so replacing T as (m1a+F) i get... m2g-(m1a+F)=m2a

that comes out to be a=(m2g-F)/(m1+m2)

so I plugged in the numbers a=(55.86-107.028)/(27.1+5.7)=-1.56m/s^2 is that right?

8. Sep 29, 2010

fizzynoob

>>T-F=m1a

where is your friction, and F has two components ... you need to break them up.

It would look something like this in the X

Forces in X = Fpcos(t) - T- Ff = (m1)a;

you can find friction by finding the normal force (sum the forces in the Y).

9. Sep 29, 2010