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Neporo4naja [7]

1 year ago

9

A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force

pulling the toy along the horizontal surface is 5N, calculate the tension in the rope

2 answers:

lana66690 [7]1 year ago

6 0

I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12

myrzilka [38]1 year ago

6 0

As per the question the rope is inclined at an angle of 60 degree with the horizontal.

Let the tension produced in the rope is T.

The effective force which pulls the toy along horizontal direction is 5 N.

Resolving T into two components we get-

$[1]\ \ Tsin\theta$ in vertical upward direction.

$[2]\ \ Tcos\theta$ in horizontal direction.

The vertical component is balanced by the weight of the body i.e

$Tsin\theta=mg$

Hence the toy moves due to the horizontal component.

$Hence\ Tcos\theta=5 N$

$Tcos60= 5N$

$T= \frac{5}{cos60}$

$= 5*2 N$ [ $cos60 = \frac{1}{2}$ ]

$= 10 N$

Hence the tension produced in the rope is 10 N.

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The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

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Using the equation of motion of the trajectory

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Using the equation of motion of the trajectory

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$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

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2 years ago

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1 year ago