A mover finds that a 20 kg crate initially at rest requires a 75 N horizontal force to set the crate in motion, but only a 60 N force to maintain a constant velocity. Find
a) the co-efficient of static friction between crate and floor.
b) the co-efficient of kinetic friction between crate and floor.
Help ASAP please!!
a. M*g = 20 * 9.8 = 196 N. = Wt. of crate = Normal(Fn).
Fap-Fs = M*a.
75-Fs = 20*0 = 0, Fs = 75 N. = Force of static friction.
us = Fs/Fn = 75/196 = 0.383.
b. 60-Fk = M*a.
60-Fk = M*0 = 0, Fk = 60 N.
uk = 60/196 = 0.306.
Sure! Let's break down the problem step by step.
a) To find the coefficient of static friction between the crate and the floor, we need to use the formula:
f_static = μ_static * N
where f_static is the static frictional force, μ_static is the coefficient of static friction, and N is the normal force.
In this case, the normal force N is equal to the weight of the crate, which can be calculated using the formula:
N = m * g
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the mass of the crate is 20 kg, we have:
N = 20 kg * 9.8 m/s^2
N = 196 N
The static frictional force, f_static, is equal to the force required to set the crate in motion, which is 75 N.
75 N = μ_static * 196 N
Now we can solve for μ_static:
μ_static = 75 N / 196 N
μ_static ≈ 0.383
Therefore, the coefficient of static friction between the crate and the floor is approximately 0.383.
b) To find the coefficient of kinetic friction between the crate and the floor, we need to use the formula:
f_kinetic = μ_kinetic * N
where f_kinetic is the kinetic frictional force, μ_kinetic is the coefficient of kinetic friction, and N is the normal force.
Using the same value of N as before (196 N), we know that the force required to maintain the constant velocity of the crate is 60 N.
60 N = μ_kinetic * 196 N
Now we can solve for μ_kinetic:
μ_kinetic = 60 N / 196 N
μ_kinetic ≈ 0.306
Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.306.
To summarize:
a) The coefficient of static friction between the crate and the floor is approximately 0.383.
b) The coefficient of kinetic friction between the crate and the floor is approximately 0.306.
To solve this problem, we can use the principles of Newton's laws of motion. Let's break down the problem step by step:
First, let's determine the static friction. The static friction is the force necessary to overcome the initial resistance and set the crate in motion. We know that the applied force to set the crate in motion is 75 N.
Therefore, the static friction force acting on the crate must be equal to the applied force, since the crate is on the verge of moving. Mathematically, we can express this as:
Fs = 75 N
Next, let's determine the kinetic friction. The kinetic friction is the force necessary to keep the crate moving at a constant velocity once it is already in motion. We know that the force required to maintain the constant velocity is 60 N.
Therefore, the kinetic friction force acting on the crate must be equal to the force required to maintain the constant velocity. Mathematically, we can express this as:
Fk = 60 N
Now, we can use the equation that relates the force of friction to the normal force and the coefficient of friction:
Ffriction = u * N
Where:
Ffriction is the force of friction,
u is the coefficient of friction, and
N is the normal force.
In this case, the normal force N is equal to the weight of the crate, which can be calculated using the formula:
N = m * g
Where:
m is the mass of the crate, and
g is the acceleration due to gravity (approximated as 9.8 m/s^2).
Given that the mass of the crate is 20 kg, we can calculate the normal force as:
N = 20 kg * 9.8 m/s^2
N = 196 N
Now we can substitute the values into the equation for static friction:
Fs = u_s * N
75 N = u_s * 196 N
Solving for u_s (the coefficient of static friction):
u_s = 75 N / 196 N
u_s ≈ 0.383
Therefore, the coefficient of static friction between the crate and the floor is approximately 0.383.
Next, we can substitute the values into the equation for kinetic friction:
Fk = u_k * N
60 N = u_k * 196 N
Solving for u_k (the coefficient of kinetic friction):
u_k = 60 N / 196 N
u_k ≈ 0.306
Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.306.
a) The coefficient of static friction between the crate and the floor is approximately 0.383.
b) The coefficient of kinetic friction between the crate and the floor is approximately 0.306.
Always remember to double-check the units and calculations to ensure accuracy.