# Incline slope help required.

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.

#### Attachments

• slope.png
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