using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity.

(a) If the force of tension in the strap is 51 N, what is the normal force on the bag?
image

(b) What is the coefficient of kinetic friction between the bag and the floor?

To solve this problem, we need to consider the forces acting on the bag and use the given information to find the normal force and the coefficient of kinetic friction.

(a) The force of tension in the strap provides the horizontal force to overcome friction and move the bag at a constant velocity. The vertical component of the tension force also supports part of the weight of the bag, reducing the normal force.

Let's break down the forces acting on the bag:

- Weight (mg): The weight of the bag is given as 15 kg. Since the bag is on a flat horizontal surface, the vertical component of the weight is balanced by the normal force.

- Tension force (Ft): The horizontal component of the tension force opposes the force of kinetic friction.

- Normal force (Fn): The normal force is the balancing force to the vertical component of the weight.

From the given information, we know that the force of tension (Ft) is 51 N, and the angle above the horizontal is 31°.

Since the force of tension is the horizontal component of the force, we can determine the vertical component using trigonometry:

Vertical component of Ft = Ft * sin(angle)
Vertical component of Ft = 51 N * sin(31°) ≈ 26.35 N

The normal force (Fn) is equal to the vertical component of the tension force plus the weight of the bag:

Fn = mg - Vertical component of Ft
Fn = (15 kg * 9.8 m/s^2) - 26.35 N
Fn ≈ 147.65 N

Therefore, the normal force on the bag is approximately 147.65 N.

(b) To find the coefficient of kinetic friction (µk) between the bag and the floor, we need to use the horizontal component of the force of tension (Ft) and the normal force (Fn).

The force of kinetic friction (Ff) can be determined using the equation:

Ff = µk * Fn

Since the bag is moving at a constant velocity, the force of kinetic friction is equal to the horizontal component of the force of tension:

Ff = Ft * cos(angle)

Substituting the given values:

Ff = 51 N * cos(31°) ≈ 43.96 N

Ff = µk * Fn

Solving for the coefficient of kinetic friction (µk):

µk = Ff / Fn
µk = 43.96 N / 147.65 N
µk ≈ 0.298

Therefore, the coefficient of kinetic friction between the bag and the floor is approximately 0.298.

To find the normal force on the bag, we need to consider the forces acting on it and analyze the vertical component of the tension force.

(a) Start by drawing a free-body diagram for the bag. We have three forces acting on the bag: its weight (mg), the tension force in the strap (T), and the normal force (N).

The weight is vertically downward with a magnitude of mg = 15 kg * 9.8 m/s^2 = 147 N.

The tension force has both a horizontal and a vertical component. The vertical component of the tension force is T * sin(31°), and it points upward.

Since the bag is moving with a constant velocity, the net force in the vertical direction must be zero. This means that the vertical component of the tension force must be equal in magnitude and opposite in direction to the weight.

So, T * sin(31°) = mg = 147 N.

Solving for T, we have T = 147 N / sin(31°) ≈ 290.23 N.

Therefore, the normal force on the bag is equal to the vertical component of the tension force, which is T * cos(31°).

N = T * cos(31°) = 290.23 N * cos(31°) ≈ 249.04 N.

The normal force on the bag is approximately 249.04 N.

(b) Now, let's find the coefficient of kinetic friction between the bag and the floor.

The horizontal component of the tension force is responsible for overcoming the force of kinetic friction.

The force of kinetic friction can be given by the equation: f_k = μ_k * N, where μ_k is the coefficient of kinetic friction.

Since the bag is moving with a constant velocity, the force of kinetic friction must be equal in magnitude and opposite in direction to the horizontal component of the tension force.

Therefore, f_k = T * cos(31°) = 290.23 N * cos(31°) ≈ 249.04 N.

Substituting this value into the equation for the force of kinetic friction, we have:

249.04 N = μ_k * 249.04 N.

Simplifying the equation, we find that the coefficient of kinetic friction is μ_k = 1.

Therefore, the coefficient of kinetic friction between the bag and the floor is approximately 1.

wouldn't it be weight minus the vertical component of the pulling force?

b) THe horizontal force=friction=mu*(mg-vertical force)=51cosTheta

bgfgt