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Lady bird [3.3K]

3 years ago

8

What force must be applied to the end of a rod along the x-axis of length 2.25 m in order to produce a torque on the rod about t

he origin of 11 z-hat N·m? Choose among the available options below. x,y,z,-z

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

<h2>4.9N approx.</h2>

Explanation:

Step one:

given data

length of rod r= 2.25m

Required torque = 11 Nm

Required

The force needed to produce a torque of 11Nm

Step two:

<em>"By definition, Torque is the twisting force that tends to cause rotation. </em>

<em>The point where the object rotates is known as the axis of rotation."</em>

Mathematically,

$\tau = rF\sin\theta$

$\tau$ = torque

r = radius

F = force

$\theta$ = angle between F and the lever arm

in this case is zero

$\tau = rF\\\\F=\tau/r$

substituting we have

F=11/2.25

F=4.88N

F=4.9N approx.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

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Generally workdone is mathematically represented as

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Answer:

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Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

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$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

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I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

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Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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