# Newton's second law (friction and pulley)

1. Mar 25, 2015

### GrantFuhrer

1. The problem statement, all variables and given/known data

2. Relevant equations
F=ma, Fr = μN

3. The attempt at a solution
1kg(1m/s^2) = 1N

Fnet = (x) - 1.18
1N = x - 1.18
x = 2.18N
2.18N/9.81m/s^2 = 0.223kg

I know this is not right because I was told that in my calculations Fnet and Fr go in the same direction. Also I didn't use the kinetic coefficient to calculate my answer (which I'm probably supposed to) the total question is out of 5 points.

2. Mar 25, 2015

### Staff: Mentor

You need to do a force balance on the blue mass also. Call T the tension in the string.

Chet

3. Mar 25, 2015

### tms

What is $F_r$ supposed to be? At one point you set it equal to the frictional force, but you also say it is in the same direction as the net force. It can't be both. Perhaps if you figure that out, all will become clear to you. I presume that the pulley is supposed to be massless and frictionless.

4. Mar 25, 2015

### GrantFuhrer

Fr is frictional force. Yes I know my mistake was that it is the same direction as the net force. Indeed the pulley is massless and frictionless. This is what I do so that the force of friction is different from the net force:
-Fnet = (x) - 1.18
-1N = (x) - 1.18
x = 0.18N
0.18N/9.81 = 0.018kg

I'm still doing it wrong, can you guide me from here?

5. Mar 25, 2015

### Staff: Mentor

T - 1.18 = 1N, but the value of T you get from this is not the weight of the blue mass. The blue mass is also accelerating, so T is less than the weight of the blue mass.

Chet

6. Mar 25, 2015

### GrantFuhrer

Hmm..

T - 1.18 = 1N
T = 2.18N / 9.81m/s^2 = 0.222kg

Same answer I had before. It's not right because when I verify:
(0.222*9.81) - 1.18 = 1 / (0.222 + 1) = 0.81

Acceleration has to be 1.00m/s^2 and not 0.81.

7. Mar 25, 2015

### tms

Just to be clear, the frictional force is not in the same direction as the net force. One thing you do know is that the tension must be greater in magnitude than the frictional force, or the block would not move. For the block on the table, you know that
$$F_{net} = T - F_{friction}$$ since the tension and the friction work in opposite directions. You had that part right in your original post. It would help if you made your equations more explicit, not using things like $x$ to represent a force.

After you get the tension, you do another free-body diagram on the hanging weight to get its mass.

8. Mar 25, 2015

mg - T = ma