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Turning effect of a force Questions
1. Figure 4 below shows a uniform metre rule in equilibrium under the forces shown
Determine the weight of the metre rule
2. The diagram below shows a uniform meter rule of mass 300g balanced by two forces F1 and F2.
Force F2 is 5N. Assuming there is no frictional force on the pulleys,
Calculate the force F1
3. (a) The figure below shows a system in equilibrium at room temperature. The system is taken
outside where the temperature is 20oC higher for sometime.
Explain why it tips to the right when it is taken outside the room.
(c) (i) State the law of floatation.
(ii) The fig. below shows a floating object of volume 40,000 cm3 and mass 10g. It is held as
shown in water of density 1.25g/cm3 by a light cable at the bottom so that ¾ of the volume
of the object is below the water surface. (Assume that up thrust due to air is negligible)
(iii) (I) Calculate the volume of the object under water. *
(II) State the volume of water displaced by the object.
(III) Calculate the weight of water displaced.
(iv) Determine the tension in the cable
(v) Calculate the density of the object.
4. State the principle of moments.
5. Figure 4 shows a uniform wooden plank which weighs 10N. The plank is balanced at 0.8m
from one end by a mass of 2.5kg
What is the length of the wooden plank in metres?
6. Figure 4 shows a uniform rod AE which is 40cm long. It has a mass of 2kg and pivoted at D. If 2N is acting at point E, and 30N force is passed through a frictionless pulley
fig 4
Find the force (x) acting at end A
7. A uniform half metre long beam, pivoted at the 10cm mark, balances when a mass of 150g is suspended at the 0cm mark as shown below:
Calculate the weight of the beam
8. The figure below shows a ring of a thin steel washer.
Determine the centre of gravity of the washer.
9. The diagram below shows a uniform metre rule balanced by two forces A and B. If force B is 5N,
assuming that there no frictional force on the fixed pulley, calculate the weight of the
metre rule.
Turning effect of a force Answers
1. Sum of clockwise moments = sum of anticlockwise moments
60cm x 200g + 50cm x Mg = 40cm x 400g *TEZ*
12000cmg + 50mcmg = 16000cmg
50Mcmg = 4000cmg
50M = 4000
M = 4000 = 80g
50
w = 0.8N
2. Calculate the force F1
40 X 5 = f X 60 + 3 X 10
200 = 60F + 30
60 F = 170
F = 170
60
= 2.833N
3. (a) Balloon and air will expand therefore up thrust on balloon increases thus clockwise moment
increases
(c) (ii) Volume under water = ¾ x 40,000
= 30,000cm3
4. For a system in equilibrium the sum of clockwise moments about a point is equal to the sum
of anticlockwise moments about the same point;
5.
0.8 x 2.5 = d x10
d = (0.8 x 25) = 2.0 (1mk)
10
length = (0.8 + 2) 2 = 5.6cm (1mk)
6. actm = ctm
x(0.3) + 2.0 x 0.1 = (30 x 0.2) + 2 x 0.1
0.3x = 6.2 – 2.0
x = 14N
7. Clockwise moments = Anticlockwise moments
1.5 X 0.1 = W X 0.15
W = 0.15/0.15
W= 1N