Question 168: Laws of Motion

Question

A mass of 10  kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $$	heta$$ = tan$$-$$1 (x $$	imes$$ 10$$-$$1). The value of x is ____________.
(Given, g = 10 m/s2)

Accepted Answer

The vertical component of the tension, $$T\cos	heta$$, balances the weight of the mass:
$$T\cos 	heta = mg$$
Since $$ mg = 10 	ext{ kg} 	imes 10 	ext{ m/s}^2 = 100 	ext{ N}$$:
$$T\cos 	heta = 100\, 	ext{N}$$ ...... (i)
The horizontal component of the tension is given by:
$$T\sin 	heta = 30\, 	ext{N}$$ ........ (ii)
Dividing equation (ii) by equation (i):
$$ \frac{T\sin 	heta}{T\cos 	heta} = \frac{30}{100}$$
Therefore:
$$ 	an 	heta = \frac{3}{10}$$
Hence, the value of x is:
$$ 	herefore x = 3$$

Laws of Motion

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