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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
Physics
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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1) 12.5 N east

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Taking east as positive direction, we can write the two forces has

F=+15 N\\F_f = -2.5 N

Therefore, the net force on the box will be:

F_{net} = F + F_f = 15 + (-2.5) = +12.5 N

And the positive sign means the direction is east.

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We can solve this part by using Newton's second law:

F_{net}=ma

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3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
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Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

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Hence\ the\ frictional\ force\ F=\mu N

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Hence the net force acting on the body having mass 2 kg-

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