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stealth61 [152]
3 years ago
15

A firefighter with a weight of 756 N slides down a vertical pole with an acceleration of 2.96 m/s2, directed downward. What are

the magnitudes and directions (choose the positive direction up) of the vertical forces
(a) on the firefighter from the pole and
(b) on the pole from the firefighter?
Physics
1 answer:
Katen [24]3 years ago
3 0

Answer:

a) F = 527.65 N, Force applied is upwards.

b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.

Explanation:

Data provided:

Weight of the firefighter = 756 N

Mass of the firefighter = 756/9.8 = 77.14 Kg

Acceleration, a = 2.96 m/s²

a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.

therefore, we have

F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.

B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)

thus, we have

F = - 527.65 N

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Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

\omega= 19 rev/s

we need to convert it into radiands per second. We know that

1 rev = 2 \pi rad

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

\theta= \omega t

where

\omega=119.3 rad

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

\theta=(119.3 rad/s)(5 s)=596.5 rad

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

\alpha = 1.6 rad/s^2

So the final angular speed will be given by

\omega_f = \omega_i + \alpha \Delta t

where

\omega_i = 119.3 rad/s is the initial angular speed

\alpha = 1.6 rad/s^2 is the angular acceleration

\Delta t = 15 s - 10 s = 5 s is the time interval

Solving the equation,

\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2

Substituting the numbers into the equation, we find

\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad

(e) 222 m

The instantaneous speed of the center of the wheel is given by

v_{CM} = \omega R (1)

where

\omega is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s

And so the displacement of the center of the wheel will be

d=v_{CM} t = (44.4 m/s)(5 s)=222 m

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dimaraw [331]
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