A woman at an airport is towing her 21.0 kg suitcase at constant speed by pulling on the strap at an angle of è above the horizontal (Fig. P4.40). She pulls on the strap with a 34.0 N force, and the friction force on the suitcase is 11.0 N.
(a) What angle does the strap make with the horizontal?
°
(b) What normal force does the ground exert on the suitcase?
N
a. Fh - Ff = ma = 0. Because a = 0.
34*cosA - 11 = 0,
34cosA = 11,
cosA = 0.324,
A = 71 Deg.
b. Fv = mg = 21kg * 9.8N/kg = 206N.
To find the angle that the strap makes with the horizontal, we can use trigonometry.
(a) The horizontal component of the force applied to the strap is the force multiplied by the cosine of the angle (θ). In this case, the force applied is 34.0 N.
Horizontal force = 34.0 N * cos(θ)
The vertical component of the force applied to the strap is the force multiplied by the sine of the angle (θ).
Vertical force = 34.0 N * sin(θ)
Since the suitcase is moving at a constant speed, the force applied to the strap must balance the friction force on the suitcase.
Friction force = 11.0 N
The vertical force also needs to balance the weight of the suitcase.
Vertical force = weight of the suitcase = mass * acceleration due to gravity
Weight of the suitcase = 21.0 kg * 9.8 m/s^2
Now, equate the vertical forces to solve for θ:
34.0 N * sin(θ) + 21.0 kg * 9.8 m/s^2= 11.0 N
34.0 N * sin(θ) = 11.0 N - 21.0 kg * 9.8 m/s^2
sin(θ) = (11.0 N - 21.0 kg * 9.8 m/s^2) / 34.0 N
Now, use the inverse sine function to find the angle θ:
θ = sin^(-1)[(11.0 N - 21.0 kg * 9.8 m/s^2) / 34.0 N]
This will give you the angle in radians. To convert to degrees, multiply by (180/π).
(b) The normal force exerted by the ground on the suitcase is the vertical force required to balance the weight of the suitcase.
Normal force = weight of the suitcase = mass * acceleration due to gravity
Normal force = 21.0 kg * 9.8 m/s^2
Calculate the value to get the answer in Newtons.