The moment of a force about a point (also called torque) is calculated as the product of the force and the perpendicular distance from the point to the line of action of the force.
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
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The moment of a force about a point (also called torque) is calculated as the product of the force and the perpendicular distance from the point to the line of action of the force.
$$ \text{Moment} = \text{Force} \times \text{Distance} $$
Step 1: Calculate the moment due to the $50 \text{ N}$ force about point O.
The force is $F_1 = 50 \text{ N}$.
The perpendicular distance from point O to the $50 \text{ N}$ force is $d_1 = 3 \text{ m}$.
$$ \text{Moment}_{50 \text{ N}} = F_1 \times d_1 = 50 \text{ N} \times 3 \text{ m} $$
$$ \text{Moment}_{50 \text{ N}} = 150 \text{ Nm} $$
Step 2: Calculate the moment due to the $80 \text{ N}$ force about point O.
The force is $F_2 = 80 \text{ N}$.
The perpendicular distance from point O to the $80 \text{ N}$ force is $d_2 = 4 \text{ m}$.
$$ \text{Moment}_{80 \text{ N}} = F_2 \times d_2 = 80 \text{ N} \times 4 \text{ m} $$
$$ \text{Moment}_{80 \text{ N}} = 320 \text{ Nm} $$
Step 3: Compare the calculated moments with the given options.
Moment of $50 \text{ N}$ is $150 \text{ Nm}$.
Moment of $80 \text{ N}$ is $320 \text{ Nm}$.
This matches option (C).
The final answer is $\boxed{\text{C) moment of } 50 \text{ N is } 150 \text{ Nm and } 80 \text{ N is } 320 \text{ Nm}}$.
Drop the next question.
