Question

Upper A 12​-pound box sits at rest on a horizontal​ surface, and there is friction between the box and the surface. One side of the surface is raised slowly to create a ramp. The friction force f opposes the direction of motion and is proportional to the normal force Upper F Subscript Upper N exerted by the surface on the box. The proportionality constant is called the coefficient of​ friction, mu. When the angle of the​ ramp, theta​, reaches 20degrees​, the box begins to slide. Find the value of mu.

Answer 1

Answer:

The coefficient of static friction is : 0.36397

Explanation:

When we have a box on a ramp of angle $\alpha$ , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.

In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

$F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)$

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

$N=m*g*cos(\alpha)$

and the force of static friction (f) is given as the static coefficient of friction ($\mu$) times the normal N:

$f=\mu *m*g*cos(\alpha)$

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

$f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)$

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

$\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397$