A dockworker loading crates on a ship finds that a 20 kg crate, initially at rest on horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60 N is required to keep it moving with a constant speed. Find the coeffiecints of static and kinetic friction between crate and floor.
Work:
75N=(20kg)a a=3.75m/s^2
60N=(20kg)a a=3m/s^2
I am not sure if I am starting it correctly. How would I solve this problem?
A dockworker loading crates on a ship finds that a 20 kg crate, initially at rest on horizontal surface requires a 75N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
No, friction= mu*m*g
75=mu*20*9.8 solve fro mu static
50=mukinetic*20*9.8 solve for mu kinetic
coefficient of friction = force app (F)/ Friction (9.8*20)
sda
To solve this problem, you need to first understand the concept of friction and the relationship between force, mass, and acceleration.
Let's break it down step by step:
1. Start with the given information:
- Mass of the crate, m = 20 kg
- Force required to set the crate in motion, Fset = 75 N
- Force required to keep the crate moving at a constant speed, Fkeep = 60 N
2. Determine the acceleration of the crate when it is set in motion:
- Use Newton's second law of motion, F = ma, where F is the net force applied to an object, m is its mass, and a is its acceleration.
- The net force that sets the crate in motion is the force applied minus the force of static friction, as the crate starts from rest.
- Therefore, Fset - fs = ma, where fs is the force of static friction.
- Substituting the given values, we have: 75 N - fs = 20 kg * a
3. Find the acceleration of the crate when it is already in motion:
- In this case, the net force applied to the crate is the force required to keep it moving, which is the force of kinetic friction.
- Therefore, Fkeep = fk = m * a, where fk is the force of kinetic friction and a is the acceleration.
- Substituting the given values, we have: 60 N = 20 kg * a
4. Solve the equations simultaneously to find the acceleration:
- From step 2, we have: 75 N - fs = 20 kg * a
- From step 3, we have: 60 N = 20 kg * a
- Equate the two expressions for a: 75 N - fs = 60 N
- Simplify, we get: fs = 15 N
5. Calculate the coefficients of friction:
- The coefficient of static friction (μs) can be found by using the formula: fs = μs * N, where N is the normal force acting on the crate (equal to its weight since it's on a horizontal surface).
- The normal force, N = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Substituting the given values, we have: 15 N = μs * (20 kg * 9.8 m/s^2)
- Simplify, we get: μs ≈ 0.077
- The coefficient of kinetic friction (μk) can be found similarly using the force of kinetic friction (fk = μk * N) since the crate is moving.
- Substituting the given values, we have: 60 N = μk * (20 kg * 9.8 m/s^2)
- Simplify, we get: μk ≈ 0.306
Therefore, the coefficients of static and kinetic friction between the crate and the floor are approximately 0.077 and 0.306, respectively.