Forces & Friction: Calculating Normal Force & Horizontal Force

[santoki]

Homework Statement

A man is pulling a 40.0 kg crate across a level floor with a horizontal force, and the coefficient of kinetic friction is μk = 0.59 for the crate and the floor.

14. What is the normal force acting on the box?
a) 231 N
b) 340 N
c) 196 N
d) 392 N

15. What horizontal force must the man apply to get the crate to accelerate at 2.0 m/s2 ?
a) 311 N
b) 231 N
c) 81 N
d) 40 N

2. The attempt at a solution

For 14:
N = mg = (40.0)(9.8) = 392N

For 15:
N-F_x = ma_x
392 - F_x = (40.0)(2.0)
F_x = 312N

but there's no 312N there and I didn't even incorporate the μk?

 

[Chestermiller]

Did you draw a free body diagram? Is N a horizontal force or a vertical force?

Chet

 

[santoki]

Did you draw a free body diagram? Is N a horizontal force or a vertical force?

Chet

N is a horizontal force.

 

[haruspex]

N is a horizontal force.

No, it's a vertical force, the vertical component of the reaction from the floor.
What, then, is the frictional force?
What is the relationship between the frictional force, the applied force, and the resulting acceleration? (Your equation was wrong.)

 

[HalfLostAndSearching]

Your calculations for the normal force and horizontal force appear to be correct. However, for question 15, you also need to take into account the frictional force, which is equal to μkN. So the equation should be N-Fx-μkN = max. Substituting the value for N, we get 392-Fx-(0.59)(392) = (40.0)(2.0). Solving for Fx gives a value of 234.8 N, which is closest to option a) 231 N. Therefore, the horizontal force the man must apply to get the crate to accelerate at 2.0 m/s2 is approximately 234.8 N.