Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. if the coefficient of static friction between a crutch and the ground is 0.913, determine the largest angle max that the crutch can have just before it begins to slip on the floor.

43 degrees

Explanation:

The force of the crutch can be broken into components.  The horizontal component of F is the static friction force keeping the crutch from sliding.  The vertical component is opposing the weight and is the Normal force.  Using the orientation of the angle q, we have the following

fs = Fx = F sin (angle (tita))

N = Fy = F cos (angle(tita))

Maximum angle implies maximum static friction

Therefore,

fsmax = UsN = Us x F cos(angle tita)

Where U = miyu

F sin(angle tita) = Us x Fcos (angle tita)

Sin (angle tita) / cos (angle tita) = Us

Therefore, tan (angle tita) = Us

Angle tita = tan^-1(Us) = tan^-1 (0.931) = 42.95 degrees = 43 degrees

There for Angle tita Max = 43 degrees

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