# Homework Help: Finding Tension and Acceleration

1. Oct 14, 2014

### B3NR4Y

1. The problem statement, all variables and given/known data
In figure, block 1 of mass m1 = 2.0 kg and block 2 of mass
m2 = 6.0 kg are connected by a massless string over a massless and
frictionless pulley and are initially held in place. Block 2 is on a rough
surface, with coefficient of kinetic friction of 0.1 and tilted at θ = 30o
while a horizontal force F of magnitude 3.0 N acts on it. Block 1 is on a frictionless surface.
Find the normal force acting on block 2.
Find the magnitude of acceleration of block 2.

2. Relevant equations
F= ma
Fg = -mg
Fg, y =-mg*cos θ
Fg, x =mg*sin θ

3. The attempt at a solution
I drew a free body diagram of each individual block, for block one I found it has a force in the positive x-direction, the force is the force of tension (therefore it will move in the direction of tension) and the force of gravity on block one is cancelled by the normal force so F T = m_ 1 * a. I solved for a, and got a = FT/m1. My reason for doing this is I reasoned the acceleration is the same for both blocks, which I may be wrong in assuming this. I solved for the x and y component of the horizontal force that acts on block 2, because my x axis runs up to down and the up direction is positive for the y axis.

To solve for the normal force, I found the y-component of the force from gravity, which is -m2 g cos 30o. I then said Y-component of gravity + Normal Force + Y-component of horizontal force = 0 and solved for Normal force. I got ~49.19 N

Then to find tension I added all the forces in the x-direction and solved for tension.

Tension + Friction Force + Force of Gravity in the X-direction + X-component of the horizontal force = m2 a
I solved for a earlier, and got FT/m1

I did the algebra and found the force of tension to be -0.31 N to be the force of tension, which has to be wrong because when I sum the forces I get a negative number.

Basically I would like to know how to solve for tension and acceleration when all you're given is the coefficient of friction, a force, the angle of the incline, and the two masses. In class we did an example where we had to solve for the tension, but we were given all this information and the acceleration of the object.

Last edited: Oct 14, 2014
2. Oct 14, 2014

### BvU

This is hard to follow without the picture.

is really a non-sequitur

It seems block 1 is on a horizontal plane ?
Is the string vertical from pulley to block 2 ?
The 3 N is working in which direction ?
How far does block 2 move when block 1 moves a small distance s ?

3. Oct 14, 2014

### B3NR4Y

Silly me, I'm stupid I knew I needed a picture... I'm on my phone currently but will post one

4. Oct 14, 2014

### B3NR4Y

Sorry friend, here it is

5. Oct 14, 2014

### Staff: Mentor

I got the same result as you for the normal force on m2. Please flesh out algebraically the force balance on m2 in the direction parallel to the incline. Writing out in terms of words is not useful to us for identifying where you encountered your problem. Thanks.

Chet

6. Oct 15, 2014

### BvU

Picture sure helps! You don't want to know what I concocted from the description only :-)
looks good, so I suspect a mishit on the calculator (*). As Chet says: show the terms (with signs!)

(The more so because I didn't get 49.2 at all: $-(-m_2g\cos{\pi\over6} + F\sin{\pi\over6}) =49.4 \$ if I use g = 9.8 m/s2 :-) )

7. Oct 15, 2014

### B3NR4Y

It must have been a mishit into my calculator or I just made up some dark energy numbers (joking), I do practice problems on a whiteboard so I don't waste paper and I got the wrong answer on the whiteboard. Today I was sitting in the library so I decided to rework the problem on paper, and I got what I assume is the right answer (I don't have the solution, it's a midterm from last year). I am not sure what I did to get the numbers I was getting :S thanks for your helps/potential helps!