A 15 kg box sits on a horizontal force.
the coefficient of static fricition between the box and the surface is 0.70. if a force P is exerted on the box at an angle directed 37 degrees below the horizontal, what must be the magnitude of P to get the box moving??
break up the force into a vertical (downward) and horizontal. The component downward adds to the force of friction.
Forcefriction=(Psin37+15*9.8)*.70
So Pcos37 must equal this force, solve for P.
To determine the magnitude of force P needed to get the box moving, we need to consider the forces acting on the box and use the concept of static friction.
The forces acting on the box include the force of gravity (mg) and the force P applied at an angle of 37 degrees below the horizontal.
Let's break down the force P into its horizontal and vertical components. The horizontal component of P is Pcos(37°), and the vertical component is Psin(37°).
Since the box is not moving initially, the static friction force (Fs) acting on the box must balance the horizontal component of force P to prevent it from moving. The formula for static friction is Fs = coefficient of static friction (μs) * normal force (N).
The normal force (N) is equal to the weight of the box, which is mg. So, N = 15 kg * 9.8 m/s^2 (acceleration due to gravity).
Using the formula for static friction, Fs = μs * N, we can calculate the static friction force.
Fs = 0.70 * (15 kg * 9.8 m/s^2) = 102.9 N
To get the box moving, we need to overcome the static friction force. Therefore, the horizontal component of force P must be greater than or equal to the static friction force.
Pcos(37°) ≥ Fs
Pcos(37°) ≥ 102.9 N
Now, we can solve for the magnitude of force P:
Pcos(37°) = 102.9 N
Divide both sides by cos(37°):
P = 102.9 N / cos(37°)
Using a calculator, evaluate the right side of the equation:
P ≈ 128.0 N
Therefore, the magnitude of force P needed to get the box moving is approximately 128.0 Newtons.