Answer :
Final answer:
The magnitude of the tension force required to keep a 12 kg block on the verge of sliding up a 30° incline with a coefficient of static friction of 0.80 is approximately 140 N.
Explanation:
To calculate the magnitude (N) of the tension force that keeps a 12 kg block on the verge of sliding up a 30° incline with a coefficient of static friction of 0.80, we need to apply Newton's second law and consider the forces along the incline. The forces acting on the block are the gravitational force component parallel to the incline (mg sin θ), the static friction force (f = μsN), and the tension force (T). The normal force (N) in this situation is equal to the gravitational force component perpendicular to the incline (mg cos θ).
First, calculate the component of the gravitational force acting along the incline:
- mg sin θ = (12 kg)(9.8 m/s²)(sin 30°) = 58.8 N.
The maximum static friction force before the block starts to slide is:
- f = μsN = 0.80(mg cos θ) = 0.80(12 kg)(9.8 m/s²)(cos 30°) = 81.7 N.
The net force (F_net) on the block is the combination of the tension force (T) and the other forces acting on the block. In this case, we are considering a scenario where the block is on the verge of sliding up; thus, the static friction force will oppose the movement, and the tension force must be equal to the sum of the gravitational force component and the static friction force.
T = mg sin θ + f
T = 58.8 N + 81.7 N = 140.5 N
Therefore, the magnitude of the tension force is approximately 140 N (option c).
