Physics, 24.07.2019 22:10 brainy51

A30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force →f so that the swing ropes are 30.0° with respect to the vertical. (a) calculate the tension in each of the two ropes supporting the swing under these conditions. (b) calculate the magnitude of →f .

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The tension in each of the two ropes supporting the girl, swing under these condition is 170 N and the magnitude of $\vec F$ is 170 N.

What is tension in the ropes due to hanging weight?

Tension is the pulling force carried by the flexible mediums like ropes, cables, and string. Tension in a body due to the weight of the suspension body is the net force acting on the body.

For the body with mass (m) hanging on a string, the tension force of the string for the component of weight in the direction of sting, can be given as,

$T=\dfrac{mg}{2\cos \theta}$

Here, (g) is the gravitational force.

(a) The tension in each of the two ropes supporting the swing under these conditions.

The weight of the girl is 30.0-kg. The girl held at rest by a horizontal force, so that the swing ropes are 30.0° with respect to the vertical. Thus, put the values in the above formula as,

$T=\dfrac{30\times9.8}{2\cos (30)}\\T=170\rm N$

Thus, the tension in each of the two ropes supporting the swing under these conditions is 170 N.

(b) The magnitude of $\vec F$ .

The tension force of the string, for the component of weight perpendicular to the sting, can be given as,

$T=\dfrac{mg}{2\sin\theta}$

In the terms of Force,

$T=\dfrac{\vec F}{2\sin\theta}$

Put the values,

$170=\dfrac{\vec F}{2\sin(30)}\\\vec F=170\rm N$

Thus, the magnitude of $\vec F$ is 170 N.

The tension in each of the two ropes supporting the girl, swing under these condition is 170 N and the magnitude of $\vec F$ is 170 N.

Learn more about the tension in the string here;

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Answer from: khaleesical9645

Part a)

T = 170 N

Part b)

F = 170 N

Explanation:

As we know that swing is at rest in this situation

so here all the forces on the swing must be balanced

so here we have

$2Tcos\theta = Mg$

so here we have

$T = \frac{Mg}{2cos\theta}$

$T = \frac{30\times 9.81}{2cos30}$

T = 170 N

Now in the horizontal direction force is also balanced

so we have

$2T sin\theta = F$

now we have

2(170)sin30 = F

F = 170 N

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A30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force →f so that the...

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