Friction and force at an angle

[SilentBlade91]

Homework Statement

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)

Homework Equations

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

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?

 

[jhae2.718]

What forces act on m₁ and m₂? You have F_g for both blocks, F_n and F_k 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.)

 

[SilentBlade91]

Well for me the block is being pushed to the left, friction to the right. But I am 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.

 

[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?

 

[SilentBlade91]

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

 

[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.

 

[SilentBlade91]

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?

 

[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).

 

[SilentBlade91]

That F value I had was after I had already subtracted the frictional force