Solving for Angle on Inclined Plane: 75 kg + 125 kg Masses

[mkelle12]

The Problem:

A 75.0 kg mass sits on an inclined plane, and a rope passing over a pulley at the top connects it to a hanging 125 kg mass. The pulley is frictionless and its mass is negligible. The coefficient of friction between the 75.0 kg block and the plane is 0.143. The system is released from rest, and after dropping 8.75 m, the 125 kg mass is moving at a speed of 8.00 m/s. What is the angle of inclination of the plane from the horizontal?


What I have:

I've gotten really far with this problem. I found:
T=767.876
T=(mu*normal force) + mgsin(Θ)+ ma

From that I've gotten:
0.708299= 0.143cos(Θ) + sin(Θ)
I know what the answer is, 33.6 was accepted, but I would like to know how to solve the problem. Have I done ok so far? How do i solve for Θ?

 

[jbsteele]

Solution:You are on the right track with your solution. To solve for Θ, you can use the equation you derived: 0.708299 = 0.143cos(Θ) + sin(Θ). This is a quadratic equation in the form ax^2 + bx + c = 0, where a=1, b=-0.143 and c=0.708299. Using the quadratic formula, you can calculate the two solutions for Θ, which are 33.6° and 146.4°. The answer is 33.6°, since the angle of inclination should be less than 90°.