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

Why did you post this question twice?

You need to be careful when you draw the force diagram. When I read the question, I drew the pulley directly above the box, but when you do that, the angle turns out to be arctan .620 = 31.8 degrees, independent of the weight.

To get the diagram right, the pulley must be placed "up the incline" so that the 45N force acts parallel to the inclined plane.

Having said that, here we go. Let the angle of inclination be x. The vertical force on the box is 130N. Resolve that into two components -- one normal (perpendicular) to the plane (Fn) and one parallel to the plane (Fp). These forces are

Fn = 130 cos x and Fp = 130 sin x

The frictional force, directed up the plane, is Fn times the coefficient of static friction 0.620. You also have the 45N force directed up the plane.

Put all this together to get the forces up and down the inclined plane:

Fp = 0.620 Fn + 45

130 sin x = 0.620 (130 cos x) + 45

130 sin x = 80.6 cos x + 45

This is not an easy equation to solve (even though we know the answer). I think we need a numeric solution using trial & error (successive approximations). To jump ahead, let's try the teacher's answer:

130 sin 49 = 80.6 cos 49 + 45
98.11 = 97.88

Close. Suppose we don't know that 49 degrees is the right answer. Let's start by defining a function

f(x) = 130 sin x - 80.6 cos x - 45

and try to find f(x) = 0. Here goes:

f(45) = (130 - 80.6) sin 45 - 45 = -10.07
(because sin 45 = cos 45)

(If I were using calculus, I'd probably take the derivative now to get my next approximation. But I won't do that since you may be taking a physics course that does not use calculus.)

For our next guess, let's try x = 50:

f(50) = 130 sin 50 - 80.6 cos 50 - 45 = 2.78

We have f(45) = -10.07 and f(50) = 2.78. To do linear interpolation, we need the equation of the line through those points. The slope is m = (2.78 + 10.07)/(50 - 45) = 2.57.

Interpolation works like this ... y - y0 = m(x - x0) ... we want the value of x when y=0, so we set y=0 and solve for x. When you do that, you get x = x0 - y0/m. Our next approximation is

x = 45 - (-10.07)/2.57 = 48.9 degrees

f(48.9) = 130 sin 48.9 - 80.6 cos 48.9 - 45 = -0.02

Wow!!! That's a better answer than 49 degrees. Go with it. To see that 48.9 is best, compare these three results:

f(48.8) = 130 sin 48.8 - 80.6 cos 48.8 - 45 = -0.28
f(48.9) = -0.02
f(49.0) = 130 sin 49 - 80.6 cos 49 - 45 = 0.23

and we were trying to get f(x) = 0. So your best answer is 48.9 degrees.

BTW, this was a good problem because of the need to get a numerical solution. We used linear interpolation, which worked fine.

Answer 2

An assumption is needed right here. i think that the forty 5 Nt is very practically adequate to pull the different weight up the slope. If i'm incorrect, and this is particularly very practically approximately to flow down the slope, then substitute the verify in Ff interior the equation under marked with ****. The putting weight provides a forty 5 Nt tension up the incline. The factor down the incline of the different container's weight is Fd. Fd = a hundred thirty Nt*sin(theta) the place theta is the attitude between the incline and horizontal. Fd resists the forty 5 Nt tension up the incline. additionally resisting this is the tension of static friction Ff. Ff = .620*N the place N is the traditional factor to the a hundred thirty Nt weight. N = a hundred thirty Nt*cos(theta). forty 5 Nt = Fd + Ff **** to evaluate: forty 5 N = a hundred thirty*sin(theta) + .620*a hundred thirty*cos(theta) could desire to unravel for theta. Hmm, i'm particular there's a miles better way than trial and blunder with distinctive values for theta, yet i don't be conscious of it. stable success with that.