Question

A 1,500-N crate is being pushed across a level floor at a constant speed by a force F of 230 N at an angle of 20.0° below the horizontal, as shown in the figure a below.

Answer 1

Step-by-step explanation:

For the vertical equilibrium of box we have

For the vertical equilibrium of box we have

N = mg + P * sin 20N = 1500 + 230sin 20 = 1579N since the crate is moving with constant speed therefore

P * cos 20 = f = mu*N * 230cos 20 = mu * 1579mu = 0.137 which is the required coefficient of friction

_(b)_

b)

Psin20 P

Pcos20 friction (f)

mg

when force P is acting up the horizontal

N + P * sin 20 = mgN = mg - Psing * 20 = 1500 - 230sir

using the second law of motion, if 'a' is the acceleration along horizontal direction we have

P * cos 20 - f = ma * 230cos 20 - 0.137 * 1421 = 1500/9.8 * a