The 50-kg wheel rolls on its hub up the circular incline under the action of the 12-kg cylinder attached to a cord around the rim. determine the angle θ at which the wheel comes to rest, assuming that friction is sufficient to prevent slippage. what is the minimum coefficient of static friction which will permit this position to be reached with no slipping?

θ=14° coefficient of static friction = 0.25

Explanation:

This is calculation on friction(body on an inclined plane)

Since the force acting parallel to the plane is the moving force(Fm)

Fm = Wsinθ (resolving the weight of the body along the horizontal)

Moving force will be mass of the cylinder × acceleration due to gravity

Fm = 12×10 = 120N

Weight of the wheel rolls = mg = 50×10 = 500N

We then have,

120 = 500sinθ

Sinθ = 120/500

θ = arcsin120/500

θ = 14°

Coefficient of friction (Ff) = uR

Where u is the coefficient of static friction and R is the normal reaction which is equal to weight(W) and Ff is equal to moving force (since object is static)

u = Ff/R = Wsinθ/ Wcosθ

u = tanθ

Since θ is 14°

u = tan 14°

u = 0.25

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radial fractures in a piece of glass look most like which of the following?

a. spokes of a wheel

b. a circle

c. a zig zag

d. none of the above

the correct answer is a. spokes of a wheel

hope this (: