High School Physics

# ICSE Physics Class 10 Force Numericals

In this post, we will solve a set of numerical problems based on force, and moment of force from the syllabus of class 10 Physics of ICSE board.

Force Numericals class 10 ICSE | Numericals on Moment of Force for class 10 ICSE

Numerical Question 1) A mechanic can open a nut by applying a force of 150 N while using a lever handle of length 40 cm. How long handle is required if he wants to open it by applying a force of only 50 N?

Solution:

In the first case, F = 150 N, and r = 40 cm = 0.4 m
Hence, moment of force = Fr = 150 x 0.4 Nm = 60 Nm
This also means that to open the nut 60 nm moment of force is required.

In the second case, F = 50 N. Required moment of force = 60 Nm.
And, lever handle length = r=?
Fr = 60
=>r =60/F = 60/50 m = 1.2 m

Numerical Question 2) A body is pivoted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. calculate the moment of force about the pivot.

Solution:

Given; F = 10 N, r = 30 cm = 0.3 m
So moment of force = Fr = 10 x 0.3 Nm = 3 Nm

Numerical Question 3) The moment of force of 5 N about a point P is 2 Nm. Calculate the distance of the point of application of the force from the point P.

Solution:

Given: the moment of force = 2 Nm, force F = 5 N.
Let, the distance of the point of application of the force from the point P = r

So, Fr = 2
=>5 r =2
r = 2/5 m = 0.4 m (answer)

Numerical Question 4) A, B, and C are the three forces each of magnitude 4N acting in the plane of paper as shown in figure. Point O lies in the same plane.

(i) Which force has the least moment about O? Give a reason.
(ii) Which force has the greatest moment about O? Give reason.
(iii) Name the forces producing (a) clockwise, and (b) anticlockwise moments.
(iv) What is the resultant torque about point O?

Solution:

(i)
We know that Moment of force = Force × Perpendicular distance

We can see from the given figure that vector C has the least perpendicular distance from point O.
Hence, vector C will have the least moment about O.

(ii) We know that Moment of force = Force × Perpendicular distance
We can see from the given figure that vector A has the greatest perpendicular distance from point O. Hence, vector A will have the greatest moment about O.

(iii)
(a) Clockwise moments are produced by vectors A and B.

Reason: If the turning effect on the body is clockwise then the moment of force is called the clockwise moment and it is taken as a negative value.

(b) Anticlockwise moment is produced by vector C.

Reason: If the turning effect on the body is anticlockwise then the moment of force is called the anticlockwise moment and it is taken as a positive value.

(iv)
Sum of torques due to vectors A, B, and C = Resultant torque about point O

= [−(4×0.9) −(4×0.8)+(4×0.6)] Nm =[−3.6−3.2+2.4] Nm=−4.4 Nm

So, the resultant torque is 4.4Nm (clockwise direction).

Numerical Question 5) The adjacent diagram shows a heavy roller, with its axle at O, which is to be raised on a pavement XY. If there is friction between the roller and pavement, show by an arrow on the diagram the point of application and the direction of force to be applied. If pivoted at 0, now will it go up?

Solution:

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Application of Force has to be done as shown in the diagram.

No, if pivoted at O it will not go up.

Numerical Question 6) A body is acted upon by two forces each of magnitude F, but in opposite directions. State the effect of the forces if
(a) both forces act at the same point of the body.
(b) the two forces act at two different points of the body at a separation r.

Solution:

(a) As two forces of the same magnitude are acting on a body at the same point and they are in opposite directions so the resultant force will be zero.

F – F = 0

(b) When two forces of the same magnitude act on a body at two different points at a separation r and in opposite direction then the moment of force will be Fr. This will tend to rotate the body about the midpoint.

[Moment of forces = F × r]

Numerical Question 7) 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:

Given,
Ab = 4 m
OA = 2 m
OB = 2 m
Force at A = 10N
Force at B = 10N

As we know, Moment of Force = F x r

Substituting the values of F and r:
The moment of the force about O at point A is 10×2 N = 20 Nm clockwise
The moment of the force about O at point B is 10×2 Nm = 20 Nm clockwise
Total moment of forces about the center O = 20+20 Nm clockwise =40 Nm clockwise

The total moment of the force about the pivot O is 40 Nm (clockwise)

Answer: Magnitude of resultant moment of forces about the pivot O = 40 Nm

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