Question

Two forces each of magnitude 5N acts vertically upwards and downwards respectively of two ends of a uniform rod of length 4m, which is pivoted at its Centre. Find the resultant moment of force about the midpoint of the rod.​

Answer 1

Answer:

Two forces each of 5N act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint .

Explanation:

Answer 2

Answer:

.•♫•♬•Given:- F1 = 5N

F2 = 5N •♬•♫•.•♬•♫•.

.•♫•♬•Length of the uniform rod :- 4m •♬•♫•.

.•♫•♬•d1 = 1/2 of 4 = 2m •♬•♫•.

.•♫•♬•d2 = 4 - d1 = 4 -2 = 2m •♬•♫•.

.•♫•♬•Now,•♬•♫•.

.•♫•♬•Formula :- T (Torque) = F × d where F is force and d is distance •♬•♫•.•♬•♫•.

.•♫•♬•Torque 1 = F1 × d1 •♬•♫•.

.•♫•♬•= 5×2 = 10m (Clockwise)•♬•♫•.

.•♫•♬•Torque 2 = F2 × d2 •♬•♫•.

.•♫•♬•= 5 × 2 = 10m (Clockwise) •♬•♫•.

.•♫•♬•Resultant Moment of force about the midpoint of the rod = Sum of Anticlockwise Moments + Sum of Clockwise Moments •♬•♫•.•♬•♫•. •♬•♫•.•♬•♫•.

.•♫•♬•= 0 + (-10 + (-10))•♬•♫•.

.•♫•♬•= 0 + (-20)•♬•♫•.

.•♫•♬•= -20m or 20m clockwise•♬•♫•.

.•♫•♬•Good morning•♬•♫•

.•♫•♬•Have a great day•♬•♫•.

.•♫•♬•Hope it helps you alot •♬•♫•.