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A system of two cylinders fixed to each other is free to rotate about a frictionless axis through the common center of the cylinders and perpendicular to the page. A rope wrapped around the cylinder of radius 2.50 m exerts a force of 4.49 N to the right on the cylinder. A rope wrapped around the cylinder of radius 1.14 m exerts a force of 9.13 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinders about the rotation axis? Answer in three decimal places.

A system of two cylinders fixed to each other is free to rotate about a frictionless axis through the common center of the cylinders and perpendicular to the pa class=

Respuesta :

The net torque is  331.402 N-m

Explanation:

Because torque is a vector quantity , the direction of torque is considered along the axis of rotation .

In first case force is applied in the right direction , thus the torque will be in upwards direction .

The magnitude of torque is = force x perpendicular distance between point of application of force and axis of rotation .

Thus in this case τ₁ = 4.49 x 2.50 = 11.25 N-m

Similarly , torque on second cylinder τ₂ = 91.3 x 1.14 = 104.08 N-m

Here force is applied downwards direction , thus the direction of torque is in the plane of page .

The torque is a vector quantity . The net torque can be found

τ = [tex]\sqrt{(11.25)^2 + (104.08)^2}[/tex] =  331.402 N-m

Torque is the force's twisting action about the axis of rotation magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.

What is torque?

Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.

The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.

In the first situation, force is delivered in the right direction, hence torque is applied upwards.

The value of torque for case 1

[tex]\rm \tau_1=4.49\times 2.50\\\\\rm \tau_1=11.25 Nm[/tex]

The value of torque for case 2

[tex]\rm \tau_2=91.3\times 1.14\\\\\rm \tau_2=104.08[/tex]

The resultant value of torque will be;

[tex]\rm \tau =\sqrt{(11.25)^2+(41.90)^2} \\\\\ \rm \tau =331.402 Nm[/tex]

Hence the magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.

To learn more about the torque refer to the link;

https://brainly.com/question/6855614

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