Question

A loading car is at rest on a track forming an angle of 25◦ with the vertical when a force is applied to the cable attached at C. The gross weight of the car and its load is 5500 lb, and it acts at point G. Knowing the tension in the cable connected at C is 5000 lb, determine (a) the acceleration of the car, (b) the distance the car moves in 20 s, (c) the time it takes for the car to return to its original position if the cable breaks after 20 s.

Answer 1

Answer:

Check attachment for solution

Explanation:

Answer 2

The motion of the car along the inclined plane reduces the force required

to pull the car upwards.

The correct values are

(a) The acceleration of the car upwards is approximately 0.09 ft./s².

(b) The distance the car moves in 20 s. is approximately 18 ft.

(c) The time it takes the car to return to its original position is approximately 1.$\underline{\overline {1}}$ seconds.

Reasons:

(a) Component of the car's weight acting along the incline plane, $W_\parallel$ = W·sin(θ)

∴ $W_\parallel$ = 5500 lbf × cos(25°) ≈ 4984.69 lbf.

The force pulling the car upwards, F = T - $W_\parallel$

Which gives;

F = 5,000 lbf - 4984.69lbf = 15.31 lbf

$Mass \ of \ the \ car = \dfrac{5500}{32.174} \approx 170.95$

The mass of the car, m ≈ 170.95 lbf

$Acceleration = \dfrac{Force}{Mass}$

$Acceleration \ of \ the \ car = \dfrac{15.31}{170.95} \approx 0.09$

The acceleration of the car, a ≈ 0.09 ft./s².

(b) The distance the car moves is given by the kinematic equation of

motion, s = u·t + 0.5·a·t², derived from Newton Laws of motion.

Where;

u = The initial velocity of the car = 0 (the car is initially at rest)

t = The time taken = 20 seconds

a = The acceleration ≈ 0.09 ft./s²

∴ s ≈ 0 × 20 + 0.5 × 0.09 × 20² = 18

The distance the car moves, s ≈ 18 ft.

(c)  If the cable breaks, we have;

Force acting downwards on the car = Weight of the car acting on the plane

∴ Force acting downwards on the car = $W_\parallel$ = 4984.69 lbf

Therefore;

$Acceleration \ of \ the \ car downwards, \ a_d = \dfrac{W_\parallel}{m}$

Which gives;

$a_d = \dfrac{4984.69 \ lbf}{170.95 \ lb} \approx 29.16 \ m/s^2$

The time it takes car to travel the 18 ft. back to its original position is given as follows;

s = u·t + 0.5·a·t²

The initial velocity, u= 0

Therefore;

$t = \mathbf{\sqrt{\dfrac{2 \cdot s}{a} }}$

Which gives;

$t = \sqrt{\dfrac{2 \times 18}{29.16} } \approx 1.\overline {1}$

The time it takes the car to return to its original position, t ≈ 1.$\overline {1}$ seconds

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