Answer:
(b) sliding down at 1.72 m/s²
(c) 14.6 N
Explanation:
A free-body diagram is a diagram showing all the forces acting on a body. In this case, there are four forces acting on the box:
- Applied force F pushing 30° above the -x axis
- Weight force mg pulling down
- Normal force N pushing right (perpendicular to the surface)
- Friction force Nμ pushing in the direction opposite of motion (where N is the normal force and μ is the coefficient of friction)
We can draw a free body diagram when the box is sliding up (friction pointing down, opposing motion) and when the box is sliding down (friction pointing up, opposing motion).
Having drawn the free body diagrams, we can now use Newton's second law of motion, which says that the net force on an object is equal to its mass times its acceleration. Applying this in the horizontal direction:
∑F = ma
N − F cos 30° = 0
N = F cos 30°
Applying in the vertical direction:
∑F = ma
F sin 30° − mg ± Nμ = ma
Substituting:
F sin 30° − mg ± F cos 30° μ = ma
(b) If the box is sliding up, then the sign of the friction force should be negative, and the sign of the acceleration should be positive. Plugging in values:
F sin 30° − mg − F cos 30° μ = ma
12 sin 30° − (1) (9.8) − 12 cos 30° (0.20) = (1) a
a = -5.88 m/s²
We got a negative answer for acceleration when we were expecting a positive one, so the box is not sliding up. It must be sliding down. Let's check by changing the sign of the friction force to positive. We should get a negative answer for acceleration.
F sin 30° − mg + F cos 30° μ = ma
12 sin 30° − (1) (9.8) + 12 cos 30° (0.20) = (1) a
a = -1.72 m/s²
Therefore, the box is sliding down at 1.72 m/s².
(c) The box is sliding down at constant speed, meaning the acceleration is zero. Solving for the applied force F:
F sin 30° − mg + F cos 30° μ = 0
F (sin 30° + μ cos 30°) − mg = 0
F (sin 30° + μ cos 30°) = mg
F (sin 30° + 0.20 cos 30°) = (1) (9.8)
F = 14.6 N
