In the following diagram, the rod is in equilibrium. Then,

$F_{1} \times d_{1} = 3 F_{2} \times d_{2}$

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$F_{1} \times d_{1} > F_{2} \times d_{2}$

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$F_{1} \times d_{1} = F_{2} \times d_{2}$

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$F_{1} \times d_{1} < F_{2} \times d_{2}$

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Solution

The correct option is C
$F_{1} \times d_{1} = F_{2} \times d_{2}$

For a body to be in equilibrium, net torque = 0 and net force = 0

Net torque = 0, i.e. sum of anticlockwise moments = sum of clockwise moments

So, we can write

$F_{1} \times d_{1} = F_{2} \times d_{2}$

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In the following diagram, the rod is in equilibrium. Then,

The correct option is C F1×d1 = F2×d2 For a body to be in equilibrium, net torque = 0 and net force = 0 Net torque = 0, i.e. sum of anticlockwise moments = sum of clockwise moments So, we can write F1×d1 = F2×d2

The correct option is C
$F_{1} \times d_{1} = F_{2} \times d_{2}$

For a body to be in equilibrium, net torque = 0 and net force = 0

Net torque = 0, i.e. sum of anticlockwise moments = sum of clockwise moments

So, we can write

$F_{1} \times d_{1} = F_{2} \times d_{2}$