Question

B) Two forces each of magnitude 5N are applied in opposite directions at the ends of a uniform rod of length 0.5 m.
Draw a diagram of the arrangement and find the total moment of the two forces.
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Answer 1

Step-by-step explanation:

hope it will be helpful to you

Answer 2

Answer:

The total moment of the two forces is 2.5Nm.

Step-by-step explanation:

Given the magnitude of forces applied on a rod,  $F_1=F_2=5N$

The length of the rod, L = 0.5 m

Moment of force or torque is given by the product of force and the perpendicular distance between a pivot and the force.

Written as $\tau=F \times r$

Diagram of the arrangement is given in the figure.

From the center of rod, the length of rod, is $\frac{L}{2} =\frac{0.5}{2}=0.25m$

Therefore, the perpendicular distance is r = 0.25m

Then, the moment of force at an end is, $\tau=F \times r$

$=5 \times 0.25=1.25Nm$

Since the same force is acting at the other end from the same distance, the moment of force at the other end is equal to 1.25Nm.

Then the resultant force is sum of the two moment of forces,

i.e., 1.25 + 1.25 = 2.5Nm  

Therefore, the total moment of the two forces is 2.5Nm.