Question

# A mass$Mkg$ is suspended by a weightless string. The horizontal force required to hold the mass at$60°$ with the vertical is

A

$Mg$

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B

$Mg\sqrt{3}$

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C

$Mg\left(\sqrt{3}+1\right)$

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D

$\frac{Mg}{\sqrt{3}}$

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Solution

## The correct option is B $Mg\sqrt{3}$Step 1. Given dataA mass$Mkg$ is suspended by a weightless stringAngle with the vertical, $=60°$Step 2. Finding the force, $F$The tension acting on the string at equilibrium is$T$ and the applied force is $F$Balancing the forcesIn the horizontal direction: $F=T\mathrm{sin}60°$ ____$\left(1\right)$In the vertical direction: $Mg=T\mathrm{cos}60°$ ____$\left(2\right)$Dividing equation $\left(1\right)by\left(2\right)$, we get: $\frac{F}{Mg}=\mathrm{tan}60°$ Therefore, $F=Mg\surd 3$Hence, option ($B$) is correct.

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