A 100-kg box is placed on a ramp. As one end of the ramp is raised, the box begins to move
downward just as the angle of inclination reaches 15◦. What is the coefficient of static friction between the box and the ramp?
Looks like one to me ...
Yep! Five consecutive posts from the same person without any thinking on her part. =(
To determine the coefficient of static friction between the box and the ramp, we need to analyze the forces acting on the box.
The forces at play here are the weight of the box (mg) acting straight downwards and the normal force (N) exerted by the ramp perpendicular to its surface.
When the ramp is inclined at an angle of 15 degrees, we can break down the gravitational force into two components: one parallel to the ramp (mg sinθ) and the other perpendicular to the ramp (mg cosθ).
In this case, the box is just about to move downward, which means it is in equilibrium. Therefore, the force of static friction (fs) must be equal to the parallel component of the weight:
fs = mg sinθ
Now, we need to find the value of fs. To do this, we'll use the equation:
fs = μsN
where μs is the coefficient of static friction and N is the normal force.
The normal force can be found using the perpendicular component of the weight:
N = mg cosθ
Substituting this back into the equation for static friction, we get:
fs = μs(mg cosθ)
Since the box is just about to move downward, the static friction force must be at its maximum value. This maximum value occurs when the static friction force is equal to the force trying to push the box downward, which is given by:
fmax = force pushing downwards = mg sinθ
Therefore, fs = fmax. Substituting this into our equation, we get:
μs(mg cosθ) = mg sinθ
Now we can solve for μs:
μs = (mg sinθ) / (mg cosθ)
Simplifying this expression, we find:
μs = tanθ
Plugging in the value of the angle of inclination (15 degrees), we can calculate the coefficient of static friction:
μs = tan(15◦)
Therefore, the coefficient of static friction between the box and the ramp is equal to the tangent of 15 degrees.