If the coefficient of kinetic friction between a 32.7 kg crate and the floor is 0.295, what horizontal force is required to move the crate at a steady speed across the floor?
(in N)
What horizontal force is required if Uk is zero?
(in N)
To get the answer, we need to use the formula for calculating the force of friction. The formula is given by:
F_friction = μ * N
Where:
F_friction is the force of friction
μ is the coefficient of kinetic friction
N is the normal force
To calculate the force of friction, we first need to calculate the normal force. The normal force is the force exerted by a surface that is perpendicular to the surface. In this case, since the crate is on the floor, the normal force would be equal to the weight of the crate.
The formula to calculate weight is given by:
Weight = mass * gravity
Where:
mass is the mass of the crate
gravity is the acceleration due to gravity (approximately 9.8 m/s^2)
For the given problem:
mass = 32.7 kg
gravity = 9.8 m/s^2
Therefore, the weight of the crate (normal force) would be:
Weight = 32.7 kg * 9.8 m/s^2
Now, we can calculate the force of friction:
F_friction = 0.295 * (32.7 kg * 9.8 m/s^2)
Simplifying the equation would give us the force of friction. However, since the problem states that the crate is moving at a steady speed, the force of friction would be equal to the force applied to move the crate at a steady speed.
Hence, the horizontal force required to move the crate at a steady speed across the floor is:
F_friction = 0.295 * (32.7 kg * 9.8 m/s^2)
To answer the second part of the question, where μ (the coefficient of kinetic friction) is zero, the force of friction would be zero as well. In this case, no horizontal force is required to move the crate at a steady speed across the floor. Therefore, the horizontal force required would be zero.