A box of books weighing 328N moves with a constant velocity across the floor when it is paged with a force of 385N exerted downward at an angle of 25.3 below the horizontal. Find muK between the box and the floor
the force increases the normal downward force.
Forcehorizontal=frictionforce=ma=0
385cos25.3=mu(328+385sin25.3)
solve for mu.
To find the coefficient of kinetic friction (μᵏ) between the box and the floor, we can use the following steps:
1. Convert the force exerted downward at an angle to its horizontal and vertical components.
- The horizontal component (Fᵥ) can be found by multiplying the force (385N) by the cosine of the angle (25.3°).
Fᵥ = 385N * cos(25.3°)
- Calculate the vertical component (Fₖ) by multiplying the force (385N) by the sine of the angle (25.3°).
Fₖ = 385N * sin(25.3°)
2. Determine the normal force (N) acting on the box.
- The normal force is equal in magnitude but opposite in direction to the vertical component of the force exerted on the box.
N = - Fₖ
(Note: We take the negative value since the normal force should act in the opposite direction of the vertical force.)
3. Calculate the net force exerted on the box.
- Since the box is moving with a constant velocity, the net force is zero.
- The net force can be determined by subtracting the force due to kinetic friction (fₖ) from the applied horizontal force (Fᵥ).
Net force = Fᵥ - fₖ = 0
4. Determine the force due to kinetic friction (fₖ).
- The force due to kinetic friction can be calculated using the equation:
fₖ = μᵏ * N
- Since the net force is zero, we can conclude that fₖ is equal in magnitude to Fᵥ.
fₖ = Fᵥ = 385N * cos(25.3°)
5. Find μᵏ by rearranging the equation from step 4:
μᵏ = fₖ / N
μᵏ = Fᵥ / N
Now we can calculate the value of μᵏ:
1. Calculate Fᵥ:
Fᵥ = 385N * cos(25.3°)
2. Calculate N:
N = - Fₖ
= - (385N * sin(25.3°))
3. Calculate μᵏ:
μᵏ = Fᵥ / N