Static Friction of box on a ramp

[elbonymcbride]

Homework Statement

A 100kg box is placed on a ramp. As part of the ramp is raised, the box begins to move downward just as the angle of inclination reaches 15 degrees. What is the coefficient of static friction between the box and the ramp?

Homework Equations

Fn = mg-Fsin15

The Attempt at a Solution

Fcos15 = Mk(mg-Fsin15)
F(cos15+sin15) = Mkmg
F=Mkmg/cos15+Mksin15
= 0.2(100kg)(9.8)\cos15+0.2sin15
I continued to solve this problem, however the answer that I got was not one of my answer choices. I believe that I used the wrong equation; but even with the right equation, I never know where to start. Can anybody help me?

 

[Doc Al]

Hint: At the point where the box just barely begins to slide, the force acting down the ramp (the component of gravity) is just overcoming the static friction.

What's the component of gravity down the ramp?
What's the maximum value of static friction?

These both depend on the angle. Set up an equation and [STRIKE]solve for the angle[/STRIKE]. Edit: You are given the angle; I meant set up an equation and solve for the coefficient of friction.

 

[elbonymcbride]

would the maximum value of static friction be zero?
How would I find the component of gravity with the information given?

 

[Doc Al]

would the maximum value of static friction be zero?

No. Hint: How does friction force relate to the normal force? How does the normal force depend on the angle?

How would I find the component of gravity with the information given?

You are given the angle of the ramp. Use a bit of trig to find the component parallel to the ramp.

 

[rwisz]

I would first review the given information and make sure all units are consistent. In this case, the mass of the box is given in kilograms and the acceleration due to gravity is given in meters per second squared, so it is important to convert the mass to kilograms.

Next, I would make sure to draw a free body diagram of the box on the ramp, including all the forces acting on it. In this case, there is the normal force (Fn), the force of gravity (mg), and the force of friction (F). The angle of inclination (15 degrees) can also be indicated on the diagram.

From the free body diagram, I can see that the only force acting in the direction of motion is the component of gravity parallel to the ramp, which is Fsin15. The normal force and the component of gravity perpendicular to the ramp (Fcos15) will cancel each other out.

Now, I can use the equation for static friction: F = μsFn, where μs is the coefficient of static friction and Fn is the normal force. From the free body diagram, I can see that Fn = mg - Fsin15. Substituting this into the equation, I get:

F = μs(mg - Fsin15)

Solving for F:

F = μsmg - μsFsin15

F + μsFsin15 = μsmg

F(1 + μssin15) = μsmg

F = μsmg / (1 + μssin15)

Now, I can plug in the given values for mass (100kg), acceleration due to gravity (9.8 m/s^2), and the angle (15 degrees) to solve for F, which is the maximum force of friction that can prevent the box from sliding down the ramp. This value can then be used to solve for the coefficient of static friction (μs) by dividing by the normal force (Fn) or by multiplying by the inverse of the normal force, which is 1/cos15.

μs = F/Fn = (μsmg / (1 + μssin15)) / (mg - Fsin15)

μs = (0.2*100*9.8 / (1 + 0.2*sin15)) / (100*9.8 - 100*9.8*sin15)

μs = 0.2 / (1 +