A crate of weight Fg is pushed by a force vector P on a horizontal floor. (a) If

Question

A crate of weight Fg is pushed by a force vector P on a horizontal floor. (a) If the coefficient of static friction is s, and vector P is directed at angle below the horizontal, show that the minimum value of P that will move the crate is given by the following equation. P = (mu_s F_g sec theta)/(1 - mu_s tan theta) (Do this on paper. Your instructor may ask you to turn in this work.) (b) Find the minimum value of P that can produce motion when s = 0.460 and Fg = 120 N for the following angles. = 0° N = 15.0° N = 30.0° N = 45.0° N = 60.0° N

Explanation / Answer

the normal force must resist both the force of gravity and the vertical compoinent of p

N = Fg + p sin theta

this moves the crate when the horizontal component of P balances of the force of frction

P cos theta = mu_s N = mu_s( F_g + P sin theta)

P cos theta = mu_s sin theta = mu_s F_g

P ( 1- mu_s tan theta) = mu_s F_g sec theta

P = (mu_s F_g sec theta)/(1 - mu_s tan theta

when = 0°

P = (0.460 ( 120 sec 0 )/(1 - (0.460) tan 0

=55.2 N

when = 15°

P = (0.460 ( 120 sec 15 )/(1 - (0.460) tan 15

=65.18 N

when = 30°

P = (0.460 ( 120 sec 30 )/(1 - (0.460) tan 30

=86.78 N

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