SOLUTIONS ON FORCE
TIME 1HR MARKS:40
SECTION A
[Attempt ALL questions in this section]
Question 1: [5 ]
[i] Moment of force acting at the pivot vertically downwards is ……….
[a] maximum
[b] minimum
[c] zero
[d] infinite.
[ii] Two or more forces acting on a circular lamina not pivoted produces…..
[a] translational motion
[b] rotational motion
[c] both translational and rotational motion
[d] none of the above
[iii] When a car moves along a ghat section the outward seeking force is….
[a] due to inertia of motion
[b] due to centrifugal force
[c] due to inertia of direction
[d] both b and c
[iv] Centre of gravity of a hit me doll is ………………
[a] towards its base
[b] at the geometric centre
[c] towards its top
[d] none of the above
[v] Principle of moments is applicable for……
[a] flexible beam
[b] non-pivoted body
[c] couple
[d] uniform beam
Question 2: [5 ]
[i] Force towards the right of fulcrum acting upwards are considered to
produce………… [ positive / negative] torque.
[ii] Centre of gravity for a boomerang lies ……. [ inside / outside] the object.
[iii] On reducing the speed of motorcycle the motorcyclist tends to come towards
the centre of the circular path due to ………. [ Centrifugal / Centripetal] force.
[iv] Longer the size of the key ……. [easier / difficult] it is to open a safe vault.
[v] Force necessary to produce uniform circular motion is ………. [ Centripetal /
Centrifugal] force.
Question 3:
[a] State two conditions for a body to be in equilibrium. [2 ]
Ans: Two conditions for a body to be in equilibrium are:
[i] algebraic sum of all the forces acting on the body is zero
[ii] algebraic sum of all the moments acting on the body is zero.
[b] Where does the centre of gravity for a hollow cone lie from its apex on its
vertical axis? [2 ]
Ans: The centre of gravity for a hollow cone lies at a height of h/3 from the
base on the vertical axis so if we must determine from its apex, it would be
at a height of [1 – 1/3] = 2/3H
[c] Two forces 20gf and 12gf acting at distance of 6cm and 10cm respectively on
either sides of the turning point of a half meter scale. Show by suitable
calculations whether the rod is in equilibrium or not. [3 ]
Ans: Applying the principle of moments,
Total clockwise moments = total anticlockwise moments.
[20 x 6] = [12 x 10]
120gfcm = 120gfcm.
Since clockwise moments = anticlockwise moments,
the half meter scale is in equilibrium.
[d] A meter scale is balanced on a knife edge at its centre. 10g weight is placed
one on top of the other at 12cm mark, the scale is found to be balanced at 45cm
mark. Calculate the mass of the meter scale. [3 ]
Ans: Let the total mass of the meter scale be M kg
Since the meter scale is uniform the weight of scale acts at 50cm mark [C.G]
Applying the Principle of moments, for the beam to be in equilibrium,
Total clockwise moments = total anticlockwise moments
M x [ 50 – 45] = 10 [ 45 – 12]
M = 66g
Hence mass of the meter scale is 0.067kg
SECTION B
[Attempt any two questions out of three]
Question 4:
[a] A uniform meter scale of mass 60g carries masses of 20g,30g and 80g from
points 10cm,20cm and 90cm marks respectively. Where must the scale be pivoted
for it to be in equilibrium? [4 ]
Ans: Let the meter scale be pivoted at X cm mark towards the left of centre.
For the meter scale to be in equilibrium applying the law of moments,
Total clockwise moments = Total anticlockwise moments
60 [ 50 – X] + 80 [ 90 – X] = 30 [ X – 20] + 20 [ X – 10]
3000 – 60X + 7200 – 80X = 30X – 600 + 20X – 200
10200 – 140X = 50X – 800
11000 = 190 X
X = 57.895 cm
The scale must be pivoted at 57.895 cm for it to be in equilibrium.
[b] Differentiate between: [i] Moment of force and Moment of Couple
[ii] Centripetal force and Centrifugal force [4 ]
Ans: [i]
Moment of force Moment of couple
Moment of force = force x Moment of couple = one
perpendicular distance of the force x couple arm
We can apply principle of The beam never is in
moments for the body to equilibrium when couple
attain equilibrium acts on a body.
[ii]
Centripetal force Centrifugal force
When a body moves in a When a body is moving in
circular path the centre a circular path the
seeking force is called outward force is called the
Centripetal force centrifugal force.
It is a real force It is a fictitious force.
[c] State the position of centre of gravity for: [2]
[i] Hollow sphere with thickness…. Geometric centre.
[ii] Solid cylinder…. Point of intersection of vertical and horizontal axes.
Question 5:
[a] State the sign conventions used for applying principle of moments when
different forces act towards the left or right of the fulcrum for the body to be in a
state of equilibrium. [4 ]
Ans: Force to the left of fulcrum acting downwards produces anticlockwise
moments taken as positive.
Force to the left of fulcrum acting upwards produces clockwise moment
taken as negative.
Force to right of the fulcrum acting downwards produces clockwise moment
taken as negative
Force to the right of fulcrum acting upwards produces anticlockwise
moment taken as positive.
[b] Where does the CG of following objects lie: [3 ]
[i] Isosceles trapezium…. Point of intersection of diagonals.
[ii] Cuboid…. Point of intersection of diagonals.
[iii] Annular ring…. Geometric centre.
[c] The following figure shows a window aircon that is attached to the wall. Due
to its weight it tends to sag outwards. To prevent it from sagging a truss member
is supporting the base of the aircon with a vertical force of 50N. Find the weight
of aircon? Given width of the cage outside the wall is 1.5m [3]
Ans: The arrangement is like a cantilever with the wall acting as the fulcrum
and point of rotation.
Also, since the cage outside is a regular cuboid, the weight would act at its
CG [ point of intersection of diagonals]. Since the weight acts vertically
downwards it would bisect the base of the aircon. i.e. at distance of 0.75m
from the wall.
Now, for the aircon to be in equilibrium, applying the principle of moments,
Total clockwise moments = Total anticlockwise moments.
W x 0.75 = 50 x 1.5
W = 100N
Hence the weight of aircon is 100N.
Question 6:
[a] Two parallel unlike forces of 10N and 25N act at 12cm from each other.
Find the point about which the body would balance. [4]
Ans: Let the force of 10N act at X cm from the fulcrum. Hence the force
of 25N would act at a distance of [12- X] cm from fulcrum on the other
side.
For the body to be in equilibrium applying the law of moments,
Total clockwise moments = Total anti-clockwise moments.
10 x X = 25 x [ 12 – X]
35X = 300
X = 8.57cm
The fulcrum should be placed at 8.57cm towards the right of 10N force
for it to be in equilibrium.
[b] A uniform meter scale is balanced at 40cm mark when weights of 20gf and
5gf are suspended at 5cm mark and 75cm mark respectively. Calculate the
weight of meter scale. [3]
Ans: Since the meter scale is uniform its weight acts at the CG i.e. 50cm
marks. For the meter scale to be in equilibrium,
Applying the principle of moments,
Total clockwise moments = Total anticlockwise moments
W [ 50 – 40] + 5 [ 75 – 40] = 20 [ 40 – 5]
10W + 175 = 700
10W = 525
W = 52.5gf
The weight of meter scale is 52.5gf.
[c] A wheel of diameter 1.6m with axle at its geometric centre is applied with
a force of 4N at point B tangential to the circumference of the wheel.
Determine the moment of force about a point A situated at its diametrically
opposite end on the periphery of the wheel. If the same force acts vertically
along the axle would the moment of force be maximum or minimum? [3]
Ans: The moment of force about point A which is at the diametrically
opposite end to point B would be = [4 x 1.6]
= 6.4Nm
If the same force acts vertically along the axle the perpendicular distance
of line of action of force would be zero hence moment of force would be
minimum equal to zero.
