Question

A crate weighing 523 N rests on a plank that makes a 22.0  angle with the ground. Find the components of the crate's weight force parallel and perpendicular to the plank. ( Fgx = –196 N, Fgy = –485 N)

Answer 1

Step-by-step explanation:

The weight of the crate will e equal to its normal reaction R

Since R = mgsin theta = W

W = mgsintheta ( resolved along the vertical perpendicular to the plank

Fy = 523sin22°

Fy = 195.92N

Since the weight is acting along the negative y direction,

Fy = -195.92N

The force parallel to the plank will be the force acting in the horizontal direction.

Fx = Fcos theta

Fx = 523cos 22°

Fx = 484.9N