NYJC
PHYSICS 9745
JC 1
Suggested solutions to tutorial on
Statics(Forces in equilibrium)
1. Forces 3 N, 4 N and 5 N are in equilibrium. Assuming that sin 37 0 = 0.6, the angle between the 3 N and 5 N forces is
370
B 530
C.900
D.1270
E.1430
Ans.
 The three forces in equilibrium must form a
closed polygon.
 The angle between any two vectors is defined
as the angle (<180o) subtended between them
at their common point of intersection from
where they are drawn and directed outwards.
 Hence, the angle between the 3 N and the 5N
forces is 1270 and NOT 530.
3N
127o
5N
A.
3N
37
o
5N
4N
37o
2. Two parallel forces of magnitudes 50 N and 80 N act in the same direction. Calculate the separation of the two forces if the
distance from the resultant force to the smaller force is 0.24 m.
[ 0.39 m ]
Ans.
50N

P
0.24 m
Fnet = 50 + 80 =130 N
d
80N



The effective net moment due to the net force about any turning point P must be equal to the
algebraic sum of the moments due to the 50 N and 80 N forces about the same turning point P.
For convenience, consider the turning point P as a point lying on the line along which the 50 N force
acts (as shown on the above diagram).
By the principle of moment , and taking moments of forces about point P
The net moment of Fnet = the algebraic sum of moments of all forces
130 N x 0.24 m = 80 N x d
 d = 0.39 m
3. Find the maximum weight W which can be supported by the structure in the
Figure 3. The maximum tension the upper rope can withstand is 1000 N and the
maximum compression the strut can withstand is 2000 N. The vertical rope is
strong enough to carry any load required. Neglect the weight of the strut.
p
300
450
[max W = 1365 N occurs when T= 1000 N and C = 1224 N ]
W
Fig. 3
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Ans.


T
Consider the equilibrium of the common junction P
which is acted upon by the tension T in the upper
rope, the tension W in the vertical rope and the
compression C in the strut meet together.
Since the junction P remains at rest, three forces are
in equilibrium, and form a closed polygon.
300
P
450
C
T
60o
30
45o
W
45o

W
o
C
Free-body diagram of point P
Using sine rule,
T / sin 450 = C / sin 600 = W / sin 750 ………(1)
If T = 1000 N (max. tension of the upper rope ), then from equation (1),

1000 / sin 450 = C / sin 600

C = (1000 sin 600 ) / sin 450

C = 1224 N ( < 2000 N ….i.e. the strut can withstand the compression. )
From equation (1) ,
1000 N / sin 450 = W / sin 750

W = (1000 sin 750 ) / sin 450 = 1365 N ( max )
However,
If C = 2000 N (max. compression of the strut ), then from equation (1),

T / sin 450 = 2000 / sin 600

T = 2000 sin 450 / sin 600

T = 1225 N ( > 1000 N ….i.e. the upper rope will snap as it cannot withstand the tension. )
4. A heavy uniform plank of length L is supported by two forces F 1 and F2 at points distant L/8 and L/4 from its ends as shown
in the Fig. 4.
L/8
L/4
What is the ratio of F1 to F2 ?
F1
Fig. 4
F2
Ans
 Since the plank remains at rest, the forces F1 , F2 and W (the gravity by the earth on the plank) are in
equilibrium.
 Net moment or torque of all the three forces about the centre of gravity the plank is zero.
 Taking moments of all force about the C.G. ,
The total clockwise moment of all forces = The total anti-clockwise moment of all forces
F1 [(1/2)L – (1/8)L] = F2 [(1/2)L – (1/4)L]
F1 [(3/8)L] = F2 [(1/4)L]
F1 / F2 = 3 / 2
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5 An underwater sonar detector floats 500 m beneath the ocean surface, attached to the upper end of a cable anchored on the
sea bed. The detector is in a watertight drum, partly air-filled for buoyancy. The drum weight is 500 N and its volume is
0.25 m3 . Assuming that the density of seawater is 1050 kg m-3 and g = 10 m s-2 , calculate
(a) the pressure on the drum due to seawater,
(b) the upthrust on the drum,
(c) the tension in the cable, assuming it is vertical.
[5.25 MPa, 2.63 kN , 2.13 k N]
Ans.
Volume (V) of drum = 0.25 m3
 = 1050 kg m-3
g = 10 m s-1
T = tension in the cable
W = gravity by the earth on the drum = 500 N
U
h= 500 m
(a) pressure on the drum by the sea = hg = 500x1050x10 = 5.25x106 Pa
(b) upthrust U = weight of the sea water displaced
= Vg = 0.25x1050x10 = 2.63x103 N
T
W
(c) at equilibrium: W + T = U,
T = U – W = 2.63x103 -500
T = 2.13x103 N
6. Th cables of a suspension bridge are anchored into large free-standing blocks of concrete as shown. One of these blocks is
shown on a larger scale. It has a length of 30 m and its cross-section and density are uniform. The maximum force which the
cables could exert on the block is 5.5 x 10 8 N for a particular bridge. The force acts in the direction shown so that its line of
action is 26 m from the point about the block might possibly rotate.
T
Fig 6(a)
Fig 6(b)
26 m
30 m
(a) Copy Fig.6(b) and show the forces which would be acting on the block if it were just about to rotate.
(b) Calculate the minimum mass of the block needed to prevent rotation when the force exerted by the cable has its
maximum value.
[ 9.7 x 107 kg ]
(c) Show on a second sketch the forces which would be acting under normal operating conditions.
Ans.
(a) The block is likely to rotate about the left edge of the base of the block at B.
At critical static equilibrium of the block i.e. at the point of being overturned, the reaction R must
be acting at the edge must be pass through the intersection of W and T.
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(b) For T = 5.5x108 N, and
T
taking moments of all forces about the edge B
T x 26 = W x 15
 W = T x 26 /15 = 5.5x108 x 26 / 15 = 9.53 x 108 N
 The minimum mass of the block required to prevent the block from being
overturned = 9.53 x 108 / 9.81 = 9.7 x107 kg
(c) Under normal operating conditions, R acts at a contact T
point X instead of the edge B since the smaller anticlock moment of T due to lesser T applied on the block
can be easily balanced by a smaller clockwise moment of
W about the contact point X
R
26 m
B
R
W
B X
30m
W
7. A step-ladder consists of two equal legs each of length 2.0 m and weight 150 N. There is no cord to help to keep the legs
together, and it rests with its legs making an angle 53 0 with the horizontal on a rough horizontal floor. What is the size of the
horizontal force which the ground must exert on each leg ?
[57 N]
Ans.
FBD of the whole ladder
FBD of the left side of the ladder
R
R
C
C
R’
N
N’
F
N
W
W
A
2m
F
W
B
F
 N and N are the normal reactions exerted on the ladder at the points of contact as shown.
 F is the friction on the ladder at the point of contact with the floor.
 R an R’ are the action and reaction pair of forces between both sides of the ladder at their hinge
 Resolving forces vertically for the whole ladder,
At equilibrium N + N’ = 2W = 2x150 = 300.
By symmetry, N = N’ , hence, N = 150 N
 Consider the FBD of the left side of the ladder, and taking moments of all forces about the common
junction C,
Total clockwise moments of forces about pt.C = Total anti-clockwise moments of forces about pt.C
N x 2cos530 = F x 2sin530 + W x 1cos530
150 x 2cos530 = F x 2sin530 + 150 x 1cos530
F = 75 / tan530 = 57 N
’
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8. In order to lift a metal wheel over a kerb of height 0.20 m, a force of 0.16 kN must be applied horizontally at the axle. The
radius of the wheel is 0.50 m. Calculate the weight of the wheel and the size of the minimum force applied at the axle that
could lift the wheel.
[ 120 N ; 96 N]
F

Ans.
N

0.50 m
A
F = 0.16 kN
0.30 m
R
0.20 m
0.40 m

W
FBD of the wheel
At static equilibrium of the wheel when it is about to be lifted over the kerb,
the normal reaction N at the contact point with the ground just turn zero
Total clockwise moments of forces about pt.A = Total anti-clockwise moments of forces about pt.A
W x 0.40 = F x 0.30
W = 0.16x103 x 0.30 / 0.40 = 120 N
Fmin
FBD of the wheel
0.50 m
A

0.30 m
R
0.20 m
0.40 m
W

For min. applied force Fmin , its moment about pt.A must be largest in order to balance the moment of
W.
 its perpendicular distance from the turning pt. A must be largest i.e. 0.50 m from pt. A
Total clockwise moments of forces about pt.A = Total anti-clockwise moments of forces about pt.A
W x 0.40 = Fmin x 0.50
W = 0.16x103 x 0.40 / 0.50 = 96 N
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9. The Figure below shows a uniform rod of mass m freely pivoted at A but kept at an angle of 45 0 to the vertical by a string
attached to the end of the rod and perpendicular to it. Find the tension in the string.
[ mg / 8 ]
A
450
Ans.

R
As the rod is in static equilibrium, the three forces W, T and R are concurrent.
 Taking moments about point A,
Assume L = length of the rod
450
T
Total clockwise moments of forces about pt.A =
Total anti-clockwise moments of forces about pt.A
A
W
450
T x L = W x (L/2) sin450
T = mg / 8
10. A ball of radius 30 mm is attached by a thread of length 20 mm to a point on a vertical frictionless wall. The ball rests in
equilibrium, touching the wall. If the weight of the ball is 8.0 N, find the push of the wall on the ball, and the pull of the
thread on the ball.
[ 6 N; 10 N ]
Ans
T
In ABC, AC = ( 30 +20 ) = 50 mm , BC = 30 mm and AB = 40 mm.
W .= 8.0 N
For static equilibrium, taking moments of all forces about A,
A
Total clockwise moments of forces about pt.A =
Total anti-clockwise moments of forces about pt.A
N
N x 40 (mm) = W x 30 ( mm )
 N = ¾ W = ¾ x 8.0 N = 6.0 N
C

B
At eqiulibrium, W, T and N form a closed polygon,
T cos  = N
T
T = N/ cos  = 6.0 N / (3/5)
= 10 N

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W
W
N
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11. A diver of weight 580 N positions herself at the end of a 4.5 m diving board of negligible weight. The board is attached by
two pedestals 1.5 m apart, as shown in the figure. Find the tension ( compression ) in each of the
two pedestals.
[ 1.16 kN ; 1.74 kN ]
1.5 m
Ans.
1.5 m
A
x1=3.0 m
T
C
R = 580 N
x2 = 4.5 m
FBD of the diving board
T = the tension acting on the board by the pedestal further from the centre of the board
C = the compression acting on the board by the pedestal nearer to the centre of the board

For static equilibrium of the diving board, taking moments of all forces about end A
Total clockwise moments of forces about pt.A = Total anti-clockwise moments of forces about pt.A
T x2 = C x1
4.5 T = 3.0 C
C = 1.5 T…………………………..(1)
Resolving forces vertically, net vertical force is zero.

From (1) and (2),
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C = T + R = T + 580 ……………….(2)
1.5 T = T + 580
0.5 T = 580
T = 1.16 kN
From (1),
C = 1.5 x 1.16 kN = 1.74 kN
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12. A beam with weight W is hinged at one end to a wall; its other end is supported by a cable wire which pulls with a force
0f 80 N at an angle of 370 with the horizontal. Find the push R of the wall on the beam. The length of the beam is 2.0 m
and its centre of gravity is 0.8 m from the wall.
Cable wire
Wall
Ans.
T
R

370
A
B
0.8 m
W
2.0 m
R = the reaction by the wall on the beam
T = tension by the cable on the beam
 As the bean is in static equilibrium, the three forces W, T and R are concurrent.
Assume L = length of the beam = 2.0 m
 Taking moments about point A,
Total clockwise moments of forces about pt.A = Total anti-clockwise moments of forces about pt.A
R x 2.0 sin = W x 0.8
R x 2.0 sin = 80 x 0.8
Rsin = 32 N …………(1)
 Taking moments about point B,
Total clockwise moments of forces about pt.B = Total anti-clockwise moments of forces about pt.B
W x 0.8 = T x 2.0 sin700
T = 80 x 0.8 / (2.0 sin700 ) = 34 N
 Resolving forces horizontally,
 FI = 0 for static equilibrium.


R cos = T cos 370 = 34 cos370 = 27 N
R cos = 27 N ………(2)
(1) / (2), 
From (2), 
tan  = 34 / 27
 = 51.50
R = 27 / cos 51.50 = 43 N
The End
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