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Leya [2.2K]
3 years ago
13

_____ friction occurs when a object rolls over a surface

Physics
2 answers:
sweet [91]3 years ago
6 0
<h2>Answer:</h2><h2><em><u>-Rolling friction occurs</u></em></h2><h3><em><u>Thank you for asking this great question need any other help please let me know by commenting below I'd be glad to help. </u></em></h3><h3><em><u>I'd also greatly appreciate you if you mark me as brainliest and click that thanks button.( optional )</u></em></h3><h3><em><u>Your brainly friend ( lauralit1 )</u></em></h3><h2><em><u /></em></h2>
Andrew [12]3 years ago
6 0

<u>Answer:</u>

<u><em>Rolling friction</em></u><em> occurs when a object rolls overs a surface.</em>

<u>Explanation:</u>

Friction is the resisting force which occurs during contact of two objects leading to preventing its motion. The loss due to friction is released in the form of heat energy.

When two objects come in contact with each other for any kind of motion, the electromagnetic attraction of the charges exhibited by the two objects leads to resisting the motion of the object with the release of heat energy.

In this case, when a object is rolling over a surface, the roughness of the surface will contribute to the friction between the surface and the rolling object leading to variation of the speed of the rolling object. This kind of friction experienced by a rolling object is termed as rolling friction.

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A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
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There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
3 years ago
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