Box weigh 130 newtons on incline and held back by 45 N box hang from frictionless pulley. Angle of incline?

Question

The 45 newton box hangs from a cord that passes over a frctionless pulley and attached to 130 newton box that is held on an incline. The answer is forty-nine degrees, but I need to show work to get credit for this problem and I do not know how to get to that answer even when my teacher gave all the students the answer. The coefficient of friction is .620.

Answer 1

You asked this question twice, and I posted a detailed solution in answer to your other question. I know your teacher's answer is 49 degrees, but if you go to the nearest tenth of a degree, 48.9 is a better answer.

Answer 2

An assumption is required here. I assume that the 45 Nt is almost enough to pull the other weight up the slope. If I'm wrong, and it's actually almost about to go down the slope, then change the sign on Ff in the equation below marked with ****.

The hanging weight provides a 45 Nt force up the incline. The component down the incline of the other box's weight is Fd.
Fd = 130 Nt*sin(theta)
where theta is the angle between the incline and horizontal. Fd resists the 45 Nt force up the incline. Also resisting that is the force of static friction Ff.
Ff = .620*N
where N is the normal component of the 130 Nt weight.
N = 130 Nt*cos(theta).

45 Nt = Fd + Ff ****

To evaluate:
45 N = 130*sin(theta) + .620*130*cos(theta)
Need to solve for theta. Hmm, I'm sure there's a better way than trial and error with different values for theta, but I don't know it. Good luck with that.

Answer 3

Hi
The system is in equilibrium, so this means that the 130N box is held back from sliding down the ramp by a force of (you guessed it) exactly the same value.

So I suppose

130xsin(t) - (45 + 130xcos(t)x0.62) = 0

where t is the angle of incline. If you substitute 49 degrees into the above equation you get near enough the right answer.
The problem is you have to solve the equation to find t.
I am sure you can do that.
The problem with all these sort of problems is setting up the right model (equation). Once that is done the maths usually isnt that hard.

regards

Answer 4

the stress (T) interior the rope ought to be equivalent to the component of the gravitational tension (m x g x sin24.0) appearing on the container so as to maintain the container from accelerating down the airplane. T = m x g x sin24.0 T = 330N x sin24.0 = 134N the traditional tension on the incline airplane is comparable to (m x g x cos24.0) n = m x g x cos24.0 n = 330N x cos24.0 = 301N wish this enables.