A 51.0 kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.710 and 0.440, respectively.
(a) What horizontal pushing force is required to just start the crate moving?
N
(b) What horizontal pushing force is required to slide the crate across the dock at a constant speed?
N
(a) What a lazy crate! It's just sitting there, not moving! To get it to start moving, we need to overcome the force of static friction. The formula to calculate static friction is given by:
f_s = μ_s * N
Where:
f_s is the force of static friction,
μ_s is the coefficient of static friction, and
N is the normal force.
The normal force is equal to the weight of the crate, so we can calculate it as:
N = m * g
Where:
m is the mass of the crate (51.0 kg), and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the numbers, we get:
N = 51.0 kg * 9.8 m/s^2 = 499.8 N
Now, plugging the coefficient of static friction (μ_s = 0.710) and the normal force (N = 499.8 N) into the formula for static friction, we get:
f_s = 0.710 * 499.8 N = 354.898 N
So, the horizontal pushing force required to just start the crate moving is approximately 354.898 N. But be careful, don't push too hard or you might send the crate flying!
(b) Alright, now the crate is in motion and we want to keep it moving at a constant speed. Since the crate is already moving, we need to consider the force of kinetic friction, which is slightly smaller than static friction.
The formula for kinetic friction is similar to the formula for static friction:
f_k = μ_k * N
Where:
f_k is the force of kinetic friction,
μ_k is the coefficient of kinetic friction, and
N is the normal force.
Again, we can calculate the normal force as:
N = m * g
Plugging in the numbers, we get:
N = 51.0 kg * 9.8 m/s^2 = 499.8 N
Now, plugging the coefficient of kinetic friction (μ_k = 0.440) and the normal force (N = 499.8 N) into the formula for kinetic friction, we get:
f_k = 0.440 * 499.8 N = 219.912 N
So, the horizontal pushing force required to slide the crate across the dock at a constant speed is approximately 219.912 N. Keep pushing, but not too hard, or you might slide right into the water!
To answer these questions, we need to understand the concepts of static and kinetic friction.
The force required to just start the crate moving is the force of static friction. The maximum static friction force can be calculated using the formula:
Fs(max) = μs * N
where Fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force.
The normal force is equal to the weight of the crate, which can be calculated as:
N = mg
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Once the crate is already moving, the force required to keep it moving at a constant speed is the force of kinetic friction. The kinetic friction force can be calculated using the formula:
Fk = μk * N
where Fk is the kinetic friction force, μk is the coefficient of kinetic friction, and N is the normal force.
Now, let's calculate the answers to the questions:
(a) To find the horizontal pushing force required to just start the crate moving, we need to calculate the maximum static friction force.
Given:
Mass of the crate (m) = 51.0 kg
Coefficient of static friction (μs) = 0.710
Acceleration due to gravity (g) ≈ 9.8 m/s^2
First, calculate the normal force:
N = mg = 51.0 kg * 9.8 m/s^2 = 499.8 N
Next, calculate the maximum static friction force:
Fs(max) = μs * N = 0.710 * 499.8 N ≈ 354.87 N
Therefore, the horizontal pushing force required to just start the crate moving is approximately 354.87 N.
(b) To find the horizontal pushing force required to slide the crate across the dock at a constant speed, we need to calculate the kinetic friction force.
Given:
Coefficient of kinetic friction (μk) = 0.440
Mass of the crate (m) = 51.0 kg
Acceleration due to gravity (g) ≈ 9.8 m/s^2
First, calculate the normal force:
N = mg = 51.0 kg * 9.8 m/s^2 = 499.8 N
Next, calculate the kinetic friction force:
Fk = μk * N = 0.440 * 499.8 N ≈ 219.91 N
Therefore, the horizontal pushing force required to slide the crate across the dock at a constant speed is approximately 219.91 N.