S 2009 H2 PHYSICS TUTORIAL: FORCES
Hooke's Law
1. A weight is suspended from a spring. The variation with weight of the length of the spring is
shown below.
length / cm
16
14
12
10
0
2
4
6
8
weight / N
What is the value of the spring constant (force constant) of the spring in N m-1?
The graph only shows the length of the spring vs the weight of the load.
To find spring constant, the force acting on the spring and the extension need to be found.
Weight of the load provides the tension in the spring to extend it.
Since F = kx
F
6
k 
 200 N m-1
x 0.15  0.12
2. The scale of a certain spring balance reads from 0 to 100 N and the scale of the spring is
0.25 m long. Find the elastic potential energy stored in the spring when the balance reads
20 N. (Assume that the spring obeys Hooke’s law.)
For a extension of 0.25 m, a force of 100 N is required.
Thus the spring constant can be obtained.
Since F = kx
F 100
k 
 400 N m-1
x 0.25
Elastic potential energy = ½ kx2
20
=) 0.050 m
400
Therefore elastic potential energy = ½ (400)(0.050)2 =0.50 J
with a force of 20 N, the extension will be (
Upthrust
3. A long narrow tube is filled with water of density 1020 kg m-3 to a depth of 1.00 m. The tube
is then inclined at 30˚ to the horizontal as shown. If atmospheric pressure is 100 kPa, what
is the pressure at point X, inside the tube?
Point X is subjected to the force of the weight above it and the force due to the atmospheric
pressure.
i.e. Px = Patm + ρfghf
where hf is the vertical height of the fluid
3
= 100 × 10 + (1020 × 9.81 × 1.0 sin 30°)
= 1.05 × 105 Pa
6. Fuel tanks often have gauges at the side which show how full the tank is (see Fig.6.1).
gauge
fuel
Fig. 6.1
Explain why the pressure at the bottom of the gauge is the same as the pressure on the
bottom of the tank.
Pressure at a point in a fuel is due to the force acting on it. In this case, the force is the weight of the
fluid above it. Even though the weight of the fluid in the tank is greater than the weight of the fluid in
the gauge, the pressure at the bottom of both side is the same. This is because pressure is the ratio
of the force over the area. The ratio of the weight over the area is the same for the tank and the
gauge.
4. An object hangs from a spring balance. The balance registers 50 N in air, 30 N when this
object is immersed in water, and 44 N when the object is immersed in another liquid of
unknown density. What is the density of the liquid?
In air
T1 = W
W = 50 N
In water
T2 + U2 = W
U2 = 50 – 30 = 20 N
In Liquid of unknown density
T3 + U3 = W
U3 = 50 – 44 = 6.0 N
Since U = ρfgVf
For submerged in water,
ρh20gVf = 20
--- (1)
For submerged in unknown liquid, ρxgVf = 6.0
--- (2)
Since the object is totally submerged in both liquid, the Vf is the same.
 X gVf
6.0
(2) / (1) =

H 20 gVf 20
X
6.0

1000 20
ρx = 300 kg m-3
5. A solid has a density of 3.5 g cm-3. Calculate the density of a liquid in which the solid would
float with one-quarter of its total volume exposed above the liquid surface.
When an object floats, it is stationary (i.e. no net force acting on it)
Weight of the object = Upthrust acting on the object
mg = fgVf
ogVo= fgVf
Since the volume of the object submerged is equal to the volume of the fluid displaced,
Vf = ¾ Vo
3.5 × g × Vo= f × g × ¾ Vo
ρf = 4.7 g cm-3
Equilibrium of coplanar forces
7. A metal disc is acted upon by a number of forces. The forces are all in the plane of the disc
and the weight of the disc is negligible. In which of the following situations is the disc in
equilibrium?
Fn = 0
τn = 0
A


B


C


D


8. A bird of weight W lands at the midpoint of a horizontal wire stretched between two poles.
The magnitude of the force exerted by each pole on the wire is F.
F
F
W
For equilibrium, Fn = 0
∑Fx = 0:
∑Fy = 0:
Since
F cos θ = F cos θ (no use to solve this ques)
F sin θ + F sin θ = W
2 F sin θ = W
F > F sin θ
2F > 2 F sin θ
2F > W
(Ans)
9. A small electrically charged sphere is suspended vertically from a thread. An oppositely
charged rod is brought close to the sphere such that the sphere is in equilibrium when
displaced from the vertical by an angle of 45°.
+
–
Which one of the following best represents the free body diagram for the sphere?
A.
B.
C.
D.
For equilibrium, the net force and net torque acting on the object must be zero.
i.e. Fn = 0 and τn = 0
Thus if the vector addition of the forces is equal to 0, i.e. the force vectors forms a closed
loop, the object is in equilibrium.
10. N98/2/1
Currents in the sea water cause the buoy to be displaced
so that the rope makes an angle of 350 with the vertical, as
shown in the above diagram. The buoy may be considered
to be acted upon by 3 forces, the tension T in the rope, a
horizontal force D and a vertical force V equal to 600 N.
(a) Using a scale of 1.0 cm to represent 100 N, complete
the figure provided to produce a vector triangle for the
three forces acting on the buoy.
(b) Use your vector triangle to determine the magnitude of
the force D.
(a)
(b) By scale drawing: D = 410 N
D
By calculation: tan35 
V
D = 600 tan 35 = 420 N
11. The diagram shows a beam of length 20.0 m and mass 40.0 kg resting on two supports
placed at 5.0 m from each end.
N2
N1
5.0 m
5.0 m
X
20.0 m
A girl of mass 50.0 kg stands on the beam between the supports. The reaction forces at the
supports are shown.
(a) State the value of N1 + N2.
(b) The girl now moves towards end X of the beam. Determine the distance of the girl from
end X when the beam is about to tip.
For equilibrium, Fn = 0 and τn = 0:
(a) Since Fn = 0  ∑Fy = 0
↑N1 + N2 = ↓ 50 × 9.81 + 40 × 9.81
N1 + N2 = 883 N
(b) When the beam is about to tip, the contact at N1 will
cease to occur, i.e. N1 = 0
Taking moment at N2,
40 × 9.81 × 5 = 50 × 9.81 × (5 – x)
x=1m
12. Rachid is using a wheelbarrow to move some blocks. When a lifting force is applied at the
handle, its support legs are lifted off the ground. The dimensions of the wheelbarrow are
shown in the diagram below.
lifting force
support leg
0.8 m
0.4 m
horizontal ground
weight = 600 N
When loaded, the total weight of the wheelbarrow and the blocks is 600 N. The ground is
horizontal.
(a) Determine
(i) the minimum vertical force needed to lift the support legs off the ground.
(ii) the magnitude and the direction of the force exerted by the ground on the wheel.
Rachid now pushes the wheelbarrow forward at constant speed. He applies a force of
260 N to the handles at an angle of 50° to the vertical.
(b) (i) Calculate the horizontal component of the force exerted by Rachid.
(ii) Determine the magnitude of the resultant frictional force acting on the wheelbarrow.
(a) (i) When the support legs are lifted from the
ground, Ns = 0
Using principle of moments with pivot at the
contact of the wheel and the ground,
FL × 1.2 = 600 × 0.4
FL = 200 N;
(ii) Since Fn = 0
↑ FL + Nw = ↓ W
Nw = 600 – 200
= 400 N ↑
(b) Since the wheelbarrow is moving at a constant
speed in a straight line, Fn = 0.
(i) Horizontal force = F sin 50°
= 260 sin 50°
= 199 N
(ii) ∑Fx = 0
f = Fhorizontal = 199 N
13. A rod of length 1.0 metre has non-uniform composition, so that the centre of gravity is not at
its geometrical centre. The rod is laid on supports across two top-pan balances as shown in
the diagram. The balances (previously set at zero) give readings of 240g and 320g.
Since the rod is stationary, it is in equilibrium, i.e. Fn = 0 and τn = 0
Taking moments about C.G (so that the moment due to the W of the rod is zero)
240 L = 320(1 – L)
L = 0.57 m
Therefore, c.g. of rod is 0.070 m to the right of the geometrical center.
14. A mountain climber is rappelling down a vertical wall (the figure below ). The rope attaches
to a buckle strapped to the climber's waist 15.0 cm to the right of his center of gravity. If the
climber weighs 770 N, find
(a) the tension in the rope and
(b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
Since the person is not moving, he is in equilibrium, i.e. Fn = 0 and τn = 0
(a) The pivot chosen is the point of contact of the wall with the climber’s feet (this is so that
the moment due to the friction and the normal force due to the wall will be 0.)
Using ∑MA = 0,
Sum of anti-clockwise moment = Sum of clockwise moment
T cos25˚ x 106 = 770 x 91
T = 729 N
(b) ∑Fx = 0
N = T sin θ
N = 729 sin 25°
= 308 N
∑Fy = 0
f + T sin θ = Wc
f = 770 – 729 cos 25°
= 109 N
R2 =3082 + 1092
R = 327 N
109
308
θ = 19.5°
R is 327 N above the horizontal.
tan 
15 A ladder of length 5.0 m and mass 2.0 kg rests against a wall. Its upper end is at a height of
4.0 m above the ground. Assuming that the wall is smooth, but not the ground, what is the
force exerted by the wall on the ladder?
Since the ladder is at rest, it is in equilibrium,
i.e. Fn = 0 and τn = 0
5.0 m
4.0 m
Take moments about bottom of ladder
(so that the moment due to X and Y is zero):
mg x 1.5 = N x 4.0
N = (2.0 x 9.81) x 1.5 ÷ 4.0 = 7.36 N
Additional:
∑Fx = 0 → X = N = 7.36 N
∑Fy = 0 → Y = mg = 2.0 x 9.81 = 19.6 N
N
X
Y
m
g