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Answer:
Question 9
State two factors affecting the turning effect of a force.
Solution 9
Moment of force about a point depends on the following two factors:
(a)The magnitude of the force applied and,
(b)The distance of line of action of the force from the axis of rotation.
Question 10
The diagram shows two forces F1= 5N and F2 = 3N acting at points A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4m
Calculate:
(i) the moment of force F1 about O.
(ii) the moment of force F2 about O.
(iii) total moment of the two forces about O.
Solution 10
Given AO=2m and OB=4m
(i)Moment of force F1(=5N) at A about the point O
=F1 x OA
=5 x 2= 10Nm (anticlockwise)
(ii)Moment of force F2(=3N) at B about the point O
= F2 x OB
=3 x 4=12 Nm(clockwise)
(iii)Total moment of forces about the mid-point O=
= 12- 10=2Nm(clockwise)
Question 11
When does a body rotate? State one way to change the direction of rotation of a body. Give a suitable example to explain your answer.
Solution 11
When the body is pivoted at a point, the force applied on the body at a suitable point rotates the body about the axis passing through the pivoted point.
The direction of rotation can be changed by changing the point of application of force. The given figure shows the anticlockwise and clockwise moments produced in a disc pivoted at its centre by changing the point of application of force F from A to B.
Question 12
Two forces each of magnitude 10N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4m which is pivoted at its mid-point O as shown. Determine the magnitude of resultant moment of forces about the pivot O.
Solution 12
Given, AB=4m hence, OA=2m and OB =2m
Moment of force F(=10N) at A about the point O
= F x OA= 10 x 2= 20Nm (clockwise)
Moment of force F (=10N) at point B about the point O
= F x OB= 10 x 2 =20Nm (clockwise)
Total moment of forces about the mid-point O=
=20 +20= 40Nm(clockwise)
Question 13
Write the expression for the moment of force about a given axis.
Solution 13
Moment of force about a given axis= Force x perpendicular distance of force from the axis of rotation.
Question 14
Figure shows two forces each of magnitude 10N acting at the points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O, situated exactly at the middle of the two forces.
Solution 14
(i)Perpendicular distance of point A from the force F=10 N at B is 0.5m , while it is zero from the force F=10N at A
Hence, moment of force about A is
= 10 N x 0.5m=5Nm(clockwise)
(ii)Perpendicular distance of point B from the force F=10 N at A is 0.5m, while it is zero from the force F=10N at B
Hence, moment of force about B is
= 10 N x 0.5m=5Nm(clockwise)
(iii)Perpendicular distance of point O from either of the forces F=10N is 0.25 m
Moment of force F(=10N) at A about O= 10N x 0.25m
=2.5Nm(clockwise)
And moment of force F(=10N) at B about O
=10N x 0.25m=2.5Nm(clockwise)
Hence, total moment of the two forces about O
=0.25 + 0.25=5Nm (clockwise)
Question 15
What do you understand by the clockwise and anticlockwise moment of force? When is it taken positive?
Solution 15
If the turning effect on the body is anticlockwise, moment of force is called anticlockwise moment and it is taken as positive while if the turning effect on the body is clockwise, moment of force is called clockwise moment and is taken negative.
Question 16
A steering wheel of diameter 0.5m is rotated anti-clockwise by applying two forces each of magnitude 5N. Draw a diagram to show the application of forces and calculate the moment of forces applied.
Solution 16
Moment of couple = either force x couple arm
= 5 N x 0.5m
=2.5 Nm
Question 17
State one way to reduce the moment of a given force about a given axis of rotation.
Solution 17
Moment of force depends on the distance of line of action of the force from the axis of rotation. Decreasing the perpendicular distance from the axis reduces the moment of a given force.
Question 18
A uniform metre rule is pivoted at its mid-point. A weight of 50gf is suspended at one end of it. Where a weight of 100gf should be suspended to keep the rule horizontal?
Solution 18
Let the 50gf weight produce anticlockwise moment about the middle point of metre rule .i.e, at 50cm.
Let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then,
according to principle of moments,
Anticlockwise moment = Clockwise moment
50gf x 50 cm=100gf x d
So, d=
Question 19
State one way to obtain a greater moment of a force about a given axis of rotation.
Solution 19
Moment of a force is the product of the force and the perpendicular distance of force from axis of rotation. So, one way to increase the moment would be to increase the distance from the axis of rotation where the force would act.
Question 20
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