Question

The dumbwaiter D and its load have a combined weight of 600 lb, while the counterweight C weighs 800 lb. Determine the power delivered by the electric motor M when the dumbwaiter (a) is moving up at a constant speed of 8 ft/s, (b) has an instantaneous velocity of 8 ft/s and an acceleration of 2.5 ft/s2, both directed upward

Answer 1

• The power delivered by electric motor M when the is moving up at a constant speed of 8 ft/s is 11,200 ft-lb/s
• When has an instantaneous velocity of 8 ft/s and an acceleration of 2.5 ft/s², the power delivered by the motor is 5,600 ft-lb/s.

Step-by-step explanation:

When the is moving up at a constant speed, its acceleration is 0. Therefore, the force acting and the counterweight is equal to the combined weight and its load (600 lb) and the counterweight (800 lb).

F = 600 lb + 800 lb = 1400 lb

The power delivered by the electric motor M can be calculated using the formula:

Power = Force × Velocity

In this case, the velocity is 8 ft/s.

Power = 1400 lb × 8 ft/s = 11,200 ft-lb/s

When an instantaneous velocity of 8 ft/s and an acceleration of 2.5 ft/s², the force acting and the counterweight can be calculated using Newton’s second law of motion.

F = 600 lb + 800 lb = 1400 lb

The acceleration of the system can be calculated using the equation:

a = (v² - u²) / 2s

where v is the instantaneous velocity (8 ft/s), u is the initial velocity (0 ft/s), and s is the distance traveled (s = 8 ft/s × 1 s = 8 ft).

a = (8² - 0²) / 2(8) = 64 / 16 = 4 ft/s²

Now, we can calculate the power delivered by the electric motor M using the formula:

Power = Force × Acceleration

Power = 1400 lb × 4 ft/s² = 5,600 ft-lb/s

Answer 2

a) 2.91 hp

b) 3.81 hp

Step-by-step explanation:

As we cans ee in the diagram given below force F has same direction as velocity
`V_D`, The power is equivalent to
`FV_D`

A) Uniform motion

As both bodies are in equilibrium,therefore we have following relation

By free body for C:

2T - 800 = 0

T = 400 lb

By free body for D

F + T - 600 lb = 0

F = 600 - T

= 600 - 400 = 200 lb

`Fv_D = 200* 8 = 1600 ft. lb/s`

`Power = 1600 * (1)/(550) = 2.91 hp`

2)

we have

`a_D = 2.5 ft/s^2   [given]`

`a_C = -(1)/(2)a_D = 1.25`

from equation of motion

by free body diagram C

F = ma

`800- 2T = (800)/(32.2)1.25`

T = 384.5 lb

by free body diagram of D

F = ma

`F + T -600 = (600)/(32.2) (2.5)`

F =262.1 lb

`FV_D = 262.1 * 8 = 2097 ft. lb/s`

`Power = (2097) (1 hp)/(550 ft. lb/s) = 3.81 hp`