# Incline slope help required.

1. Sep 8, 2013

### lecammm

As given by the attached question, we are given:
1. The force on the pulley, 51N
2. The angle of the pulley, 31 degrees
3. The mass of the box, 15 kg
4. The angle of the incline, 31 degrees
5. The kinetic friction coefficient, 0.5

I think the relevant equations are:

ƩFy = 0;
ƩFx = ma;
F(friction) = N*μ

My attempt of the solution involved choosing my axis to be parallel to the slope (x axis) and y axis perpendicular to the slope, then breaking the 51N force on the pulley up into it's components, i.e;

for x direction: 51cos(31)
for y direction: 51sin(31)

Then break up the boxes weight force into x & y components, i.e.

for x direction: (15*9.81)*cos(31)
for y direction: (15*9.81)*sin(31)

Next I summed the forces in the y direction to find the normal force N

i.e.

ƩFy = -(15*9.81*sin(31))+N+51(sin(31) = 0

Giving

N = 49.521

Next I found the force for friction giving:

F = 49.521*0.5
F = 24.7605

So then I tried to calculate the forces in the x direction with the positive direction being up the slope

ƩFx = 24.7605 + 51cos(31) + T - (15*9.81)*cos(31) + F = 15 * a

However I do not know where to find T, or if any of this working is correct, I'd really like some guidance on where to go from here, thanks.

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2. Sep 8, 2013

### CWatters

Are you sure the force on the pulley 51N ? The diagram shows the tension in the rope is 51N.

The tension in the rope will be the same both sides of the pulley. One side is parallel with the slope. One side at an angle. The total component parallel with the slope will be more than 51N.

You haven't explained what T is (for example it's not shown on the diagram).

Last edited: Sep 8, 2013