# 2 masses 1 on incline

## Homework Statement

A mass m2 = 1.1 kg, on a 35.0° incline, is connected to a mass m1 = 6.0 kg, on a horizontal surface. The surfaces and the pulley are frictionless. If F = 17.5 N, what is the magnitude of the tension in the connecting cord?
http://img717.imageshack.us/img717/1943/hmwork1.gif [Broken]

F=ma

## The Attempt at a Solution

Realy stumped on this one, i drew free body diagrams for each of the objects and attempted to eliminate the forces, i believe that F-(tension+m*g*cos(35)) = net force. Then i add the masses together and devide by M to get acceleration, and then plug acceleration back into f=ma but instead of the total mass i use 1.1 for m2.

Last edited by a moderator:

rl.bhat
Homework Helper
Hi juggalomike, welcome to PF.
F-(tension+m*g*cos(35)) = net force
This step should be
F-(tension+m*g*sin(35)) = m2*a ......(1)
Next T = m1*a...(2)
Solve these two equation to find a and T.

i dont understand what mgsin is... are you talking about m1 or m2? help i am trying to figure this problem out and i still dont get it...

mgsinθ is the resultant of normal force and gravity on the mass on the ramp.

i still get it wrong.. i dont understand how to get "a" when tension is in the eq.... F-(tension+m*g*sin(35)) = m2*a ......(1) :( helppp

You can us $ma$ in place of tension. Both objects will be accelerating at the same rate, so both a's will be the same. You now only have 1 variable, and can solve for acceleration.

thank you for trying it seems i am doing something wrong bc it tells me its wrong. :(

Well, tell us the answer, tell us what you are getting and show your work. I'm sure that we can help you figure out what you are doing wrong.

wait! i was doing a silly mistake!! i got it! thank you thank you thank you!!! :)

I still don't understand how to solve this problem.. please explain?