A woman at an airport is towing her 20.0 kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal. She pulls on the strap with a 35.0 N force, and the friction force on the suitcase is 20.0 N.
a) What angle does the strap make with the horizontal?
b) What normal force does the ground exert on the suitcase?
(a) θ = cos^-1 (20/35) = 55 degrees
(Friction equals the horizontal component of the applied force, since it is not accelerating.)
(b) M g - 35 sin 55 = 196 - 28.7 = 167.3 N
sixota
To solve this problem, we'll need to break down the forces and use Newton's second law.
a) To find the angle the strap makes with the horizontal, we can use the fact that the woman is towing the suitcase at a constant speed. This means that the net force acting on the suitcase must be zero, so the force of friction must balance the force applied by the woman.
Let's break down the forces acting on the suitcase:
1. The woman is applying a force of 35.0 N at an angle θ above the horizontal.
2. The friction force on the suitcase is 20.0 N, acting opposite to the direction of motion.
Since the suitcase is moving at a constant speed, the net force acting on it is zero. This means that the horizontal component of the applied force is equal to the friction force:
Horizontal component of applied force = Friction force
35.0 N * cos(θ) = 20.0 N
Rearranging the equation, we have:
cos(θ) = 20.0 N / 35.0 N
cos(θ) ≈ 0.5714
Taking the inverse cosine (arccos) of both sides:
θ ≈ arccos(0.5714)
θ ≈ 55.24 degrees
Therefore, the angle the strap makes with the horizontal is approximately 55.24 degrees.
b) To find the normal force exerted by the ground on the suitcase, we need to consider the vertical forces acting on it.
1. The weight of the suitcase, which is equal to its mass (20.0 kg) multiplied by the acceleration due to gravity (9.8 m/s^2): Weight = 20.0 kg * 9.8 m/s^2 = 196.0 N
2. The vertical component of the applied force, which is equal to the applied force (35.0 N) multiplied by the sine of the angle θ: Vertical component of applied force = 35.0 N * sin(θ)
Since the suitcase is not accelerating vertically, the normal force exerted by the ground must balance the vertical forces:
Normal force = Weight + Vertical component of applied force
Normal force = 196.0 N + 35.0 N * sin(θ)
Plugging in the value of θ we calculated earlier:
Normal force ≈ 196.0 N + 35.0 N * sin(55.24 degrees)
Calculating the value:
Normal force ≈ 196.0 N + 35.0 N * 0.8192
Normal force ≈ 196.0 N + 28.67 N
Normal force ≈ 224.67 N
Therefore, the normal force exerted by the ground on the suitcase is approximately 224.67 N.
To solve this problem, we need to analyze the forces acting on the suitcase.
a) First, let's consider the forces in the horizontal direction. The only force acting in the horizontal direction is the force applied by the woman, which has a magnitude of 35.0 N. Since the suitcase is moving at a constant speed, the net force in the horizontal direction must be zero. This means the friction force, which is in the opposite direction of the applied force, must be equal to 35.0 N.
b) Now, let's consider the forces in the vertical direction. Since the suitcase is not accelerating vertically, the net force in the vertical direction must be zero as well. The vertical forces acting on the suitcase are the weight (mg) and the normal force (N) exerted by the ground on the suitcase.
Given that the mass of the suitcase is 20.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2, the weight can be calculated as follows:
weight = mass * acceleration due to gravity
weight = 20.0 kg * 9.8 m/s^2
Using the given information that the friction force is 20.0 N, we can conclude that the normal force must also be 20.0 N.
a) To find the angle θ that the strap makes with the horizontal, we can construct a right triangle using the vertical (weight) and horizontal (friction) forces. By using the trigonometric function tangent, we can find the angle:
tangent(θ) = vertical component / horizontal component
tangent(θ) = weight / friction
Substituting the known values, we have:
tangent(θ) = (mass * acceleration due to gravity) / friction
θ = arctangent((mass * acceleration due to gravity) / friction)
Plug in the values for the mass, acceleration due to gravity, and friction to calculate the angle θ.