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gogolik [260]
1 year ago
8

An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a lengt

h of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)

Physics
1 answer:
ehidna [41]1 year ago
3 0

The the tension in the horizontal rope is 7,019.4 N.

<h3>What is tension?</h3>

Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.

Tension can also be described as the action-reaction pair of forces acting at each end of said elements.

The unit of tension is Newton (same unit as force, since tension is also a force).

<h3>Tension in the horizontal rope</h3>

The tension in the horizontal rope is calculated as follows;

Apply the principle of torque;

T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ

T = (M + 2m)g sinθ/cosθ

T =  (M + 2m)g tanθ

<em>let θ = 66⁰ (this value should be given in the question)</em>

<em />

T =  (M + 2m)g tanθ

T = (74.9    +   2 x 122)(9.8)(tan 66)

T = 7,019.4 N

Thus, the the tension in the horizontal rope is 7,019.4 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

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Answer:

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In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

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T=2\pi \sqrt{\frac{l}{g}}         (1)

l: length of the pendulum = ?

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T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

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Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

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  • Two intersections: The planet blocks the camera's view of the top of the building.
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There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle \angle \mathrm{B\hat{C}D} which corresponds to this minor arc.

This angle comes can be split into two parts:

\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}.

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\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}.

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\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}.

The lengths of segment DC, AC, BC can all be found:

  • \rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m;
  • \rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m;
  • \rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m.

In the two right triangles \triangle\mathrm{DAC} and \triangle \rm BAC, the value of \angle \mathrm{B\hat{C}A} and \angle \mathrm{A\hat{C}D} can be found using the inverse cosine function:

\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}

\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}

\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}.

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