A mass of 10 kg is suspended from a spring b | vedclass

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(A) Let $T$ be the tension in the spring balance. The forces acting on the mass $M$ are the tension $T$ along the string,the gravitational force $mg$

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$20 \ kg-wt$

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(A) Let $T$ be the tension in the spring balance. The forces acting on the mass $M$ are the tension $T$ along the string,the gravitational force $mg$ acting vertically downwards,and the horizontal force $F$ applied by the string.In equilibrium,the vertical component of the tension $T$ must balance the weight of the mass.$T \cos 	heta = mg$Given $m = 10 \ kg$,$	heta = 60^{\circ}$,and $g$ is the acceleration due to gravity.The reading of the spring balance corresponds to the tension $T$ in $kg-wt$.$T = \frac{mg}{\cos 60^{\circ}}$Since $1 \ kg-wt = 1 \ kg 	imes g$,the reading in $kg-wt$ is simply $T/g = m / \cos 60^{\circ}$.$T_{reading} = \frac{10}{\cos 60^{\circ}} = \frac{10}{1/2} = 20 \ kg-wt$.

$A$ mass of $10 \ kg$ is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of $60^{\circ}$ with the vertical. The new reading of the balance is

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