# Friction and tension problem

1. Oct 10, 2008

### Sagekilla

1. The problem statement, all variables and given/known data
Find the the acceleration, a[/i, and the Tension, T, in the system shown below if (a) m1 = 5 kg and (b) m1 = 2 kg. assume the coefficient of kinetic fricton on the incline plane is µk = 0.1.

Other relevant information:
Mass 1 is a 5 or 2 kg block on a 30 degree incline plane (based so that Fy is Fsin(30) ) connected to a massless rope that holds Mass 2, a 5 kg block, over a pulley.

2. Relevant equations
Ff = (µk)(Fn)
w = mg (acceleration due to gravity is assumed to be 10 m/s^2 for easy calculation according to the professor)
Fnet = ma

3. The attempt at a solution

I solved for the weight of mass 1 as being:

m1 = 5 kg: W = 50 N * sin(30) = 25 N
m2 = 2 kg: W = 20 N * sin(30) = 10 N

And found force of friction:
Ff = 0.1 * 25 N = 2.5 N

The weight of block m2:
w = 10 m/s^2 * 5 kg = 50 N

I tried creating a FBD for block m1, where the following forces were applied:
W = 25 N
Fn = 25 N
Ff = 2.5 N
F = Wm2 (Wm2 = 50 N)

And a FBD for block m2, where:
W = 50 N

Except, I'm confused where to go from here now. I know I have to calculate tension still. I'm thinking that the total force being applied on m1 according to Newton's 2nd Law is:
Fnet = ma - Ff = 50 N - 2.5 N = 47.5 N

And acceleration is: a = 47.5 N / 5 kg = 9.5 m/s^2

2nd attempt:
I know Fnet = ma - T. and Fnet is not 0 N, it's 50 N. Since the rope is massless the tension is 2T = 100 N.

I therefore arrive at the following answers of:
a = 9.5 m/s^2
T = 100 N

Does this look right?

Last edited: Oct 10, 2008
2. Oct 10, 2008

### Redbelly98

Staff Emeritus
Welcome to PF, Sagekilla.

Do you mean the block is 30 degrees from the vertical direction?

I see some issues here, mainly with terminology.

How about calling these mA and mB instead, since you say earlier that Mass 2 is fixed at 5 kg. It would make sense then to let m2 be the mass of Mass 2.

Also, the weight is simply m times g. It looks like you're calculating the weight force's y-component (because of the trig terms). Is this supposed to be the normal force--I see you use these numbers for calculating friction force next?

True if 25 N is the normal force, which is true if 30 degrees is with respect to vertical.

Okay, so now m2 is for Mass 2?

Again:

W = m1 g = ____?
Fn -- looks okay.
Ff -- looks okay.
F (rope tension?) is not necessarily the weight of Mass 2, since there is an acceleration.

The rope tension should appear here too.

Not quite.
Fnet = ma, period

I'm gonna stop here for now, and give you a chance to work through these comments before going further.

3. Oct 10, 2008

### Sagekilla

I did the problem once more, this time only trying to solve for (a) to make things less confusing:

I found the Force in the x direction for mass 1:
F = 5 kg * 10 m/s^2 * cos(30) = 43.3 N

I solved for weight of mass m1:
w1 = 5 kg * 10 m/s^2 * sin(30) = 25 N

The normal force is the same magnitude as the weight for mass m1:
Fn = w1 = 25 N

And solved for the Frictional force, Ff, with coefficient of friction 0.1:
Ff = 0.1 * 25 N = 2.5 N

I also solved for weight of mass m2:
w2 = 5 kg * 10 m/s^2 = 50 N

w = -25 N
Fn = 25 N
Ff = -2.5 N
F = -43.3 N
T = ?

The FBD for m2 looks like this:
w = -50 N
T = ?

Last edited: Oct 10, 2008
4. Oct 11, 2008

### Redbelly98

Staff Emeritus