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musickatia [10]
3 years ago
14

If a mass of 76 kg acts downward 0.38 m from the axis of rotation on one end of a board and another force of 129 N also acts dow

nward, what is the moment arm of the second force to balance this system
Physics
1 answer:
Semenov [28]3 years ago
6 0

Answer:

y = 2.196 m

Explanation:

Mass, m = 76 kg

distance from axis of rotation, x = 0.38 m

Second Force, F = 129 N

moment arm of the second force, y = ?

Now, equating moments for the equilibrium

So,

m g × x = F x y

76 x 0.38 x 9.81 = 129 x y

y = \dfrac{76\times 0.38\times 9.81}{129}

y = 2.196 m

Hence, the length of the moment arm is equal to 2.196 m.

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The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration​
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Torque from the rotational movement is defined as

\tau = I\alpha

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The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

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\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

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2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

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With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

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\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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3 years ago
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