1. A door is 3 feet wide. Calculate the moment of force when a force of 100 N is applied at
a. the end of the door
b. the middle of the door
c. 1 feet away from the hinge
Solution:
a.Force of 100 N applied at the end of the door.
Convert 3 feet into SI unit of metre.
1 ft = 30 cm.
3 ft = 90 cm = 0.9 m
Moment of force = force x perpendicular distance
= 100 x 0.9 = 90 Nm
b. Force of 100 N applied at the middle of the door
1.5 ft = 1.5 x 30 = 45 cm = 0.45 m
Moment of force = 100 x 0.45 = 45 Nm
c. Force of 100 N applied 1 ft away from the hinge (pivot)
1 ft = 30 cm = 0.3 m
Moment of force = 100 x 0.3 = 30 Nm
We see that the lower the distance from the fixed point, the lower is the torque or moment of force. For the force applied as given in the picture, the door rotates in the anticlockwise direction.
2. Three forces A, B, C act around a fixed point O. What of the total torque, and in what direction is the resultant torque? All three forces are 10 N each.
Solution:
OA = 25 cm = 0.25 m
OB = 20 cm = 0.2 m
OC = 40 cm = 0.4 m
Torque due to B, TB = 10 x 0.20 = 2.0 Nm, anticlockwise direction
Torque due to C, TC = 10 x 0.40 = 4.0 Nm, clockwise direction
Anticlockwise moment is considered positive.
Clockwise moment is considered negative.
Total torque is obtained as – TA – TC + TB = – 2.5 – 4 + 2 = – 4.5 Nm in the clockwise direction, as it is negative.
3. Two forces are equal and opposite at 2 N, and separated by a distance of 1 m. Find the moment of force -
a. exactly in the middle
b. on one of the force
Solution:
Moment due to X, TX = 2 x 0.5 = 1 Nm, clockwise direction
Moment due to Y, TY = 2 x 0.5 = 1 Nm, clockwise direction
Total moment is –TX – TY = -1 – 1 = -2 Nm (clockwise direction)
Moment due to X, TX = 2 x 0 =0 Nm (Point ‘b’ is on the line of force X, so its distance from X is zero)
Moment due to Y, TY = 2 x 1 = 2 Nm, clockwise direction
Total moment is TX – TY = 0 – 2 = -2 Nm (clockwise direction)
4. A heavy drum is to be raised a step. Where must the least force be applied, at X or Y?
Solution:
Moment of force = force x perpendicular distance
To raise the drum towards the right, it has to be rotated clockwise. The drum must roll over point A which is the fixed point. For maximum moment with minimum force, distance must be largest.
Case i: Tangential force at Y
The perpendicular distance between the line of action of force and pivot is AZ, which is the radius of the drum.
Case ii: Tangential force at X
The perpendicular distance between the line of action of force and pivot is AX, which is the diameter of the drum.
Since diameter > radius, the least force can be applied at X to roll up the drum.
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