A cube of mass M rests tilted against a wall as shown. There is no friction betw

Question

A cube of mass M rests tilted against a wall as shown. There is no friction between the wall and the cube, but the friction between the cube and the floor is just sufficient to keep the cube from slipping. When Theta is between 0 and 45 degrees, find the minimum coefficient of static friction as a fuction of Theta.

A cube of mass M rests tilted against a wall as shown. There is no friction between the wall and the cube, but the friction between the cube and the floor is just sufficient to keep the cube from slipping. When Theta is between 0 and 45 degrees, find the minimum coefficient of static friction as a function of Theta.

Explanation / Answer

let normal force on the cube by vertical wall = N

normal force on cube by ground = Mg

so fricitonal force = uMg

equating horizontal forces

N = uMg

Torque about origin = 0

so Mg*x(1-cos(45+theta))= N*x*sin theta

where x is side of cube

substitute N =uMg here

so Mg*x(1-cos(45+theta))= uMg*x*sin theta

so u = (1-cos(45+theta)/sin theta = (sqrt(2) - cos theta + sin theta)/sqrt(2)*sin theta

for u to be munimum differentiate u with respect to theta, and equate to 0

du/d theta = 2cos^2 theta + sqrt(2) cos theta - 1 = 0

so we get cos theta = 0.437

so substituting that theta in u we get minimum value of coefficient of static friction u

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