If the coefficient of kinetic friction between a 36 kg crate and the floor is 0.31, what horizontal force is required to move the crate at a constant velocity (so that FNET=0.0 N) across the floor?
m g mu = 36 * 9.81 * .31
To find the horizontal force required to move the crate at a constant velocity across the floor, we need to consider the forces acting on the crate.
1. The force of gravity acts vertically downward and is given by:
F_gravity = m * g
where m is the mass of the crate (36 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The normal force exerted by the floor on the crate cancels out the vertical component of the gravitational force. So, the normal force is equal to the weight of the crate:
F_normal = F_gravity = m * g
3. The frictional force opposes the motion of the crate and is given by:
F_friction = μ * F_normal
where μ is the coefficient of kinetic friction (0.31) and F_normal is the normal force.
Since the crate is moving at a constant velocity, the net force acting on it must be zero. Therefore, the force of friction must be equal and opposite to the applied force:
F_applied = F_friction
Substituting the values into the equations:
F_friction = (0.31) * (m * g)
F_applied = (0.31) * (m * g)
Now, we can calculate the horizontal force required to move the crate:
F_applied = (0.31) * (36 kg * 9.8 m/s^2)
F_applied ≈ 108.26 N
The horizontal force required to move the crate at a constant velocity across the floor is approximately 108.26 N.