Questions
 The figure below shows a uniform bar of length 1 m pivoted near one end. The bar is kept in equilibrium by a spring balance as shown.
Given that the reading of the spring balance is 0.6N. Determine the weight of the bar.  The figure shows a device for closing a steam outlet.
The area of the piston is 4.0 x 10^{4} m^{2} and the pressure of the steam in the boiler is 2.0 x 10^{5 }Nm^{2}. Determine the weight W that just holds the bar in the horizontal position shown.  The diagram below shows a uniform bar of lengths 6m. If the weight of the bar is 15N, determine x.
 State the principle of moments
 Name four activities which produce a turning effect
 Why is it very difficult to open a door from a point too close to hinges
 Why are people who are maimed or have lost one leg provided with crutches?
 A uniform half metre rod is balanced by a weight of 38N at one end. If the pivot is placed 10cm from the same end, calculate the weight of the rod.
 Two forces of 10N and 20N when applied at ends A and B respectively are just able to lift a non uniform rod of lengths 2m.
 What is the weight of the rod?
 Determine the position of the centre of gravity of the rod
 Determine the value of Y in the diagram below
 The figure below shows force F1 and F2 acting on a metre rule such that it is in equilibrium.
Mark on the figure a third force F3 acting on the rule such that the equilibrium is maintained. 
 State the principle of moments.
 Two men P and Q carried a uniform ladder 3.6 m long weighing 1200N. P held the ladder from one end while Q supported the ladder at a point 0.4m from the other end.
 Sketch a diagram showing the forces acting on the ladder.
 Calculate the load supported by each man.
 The figure shows a uniform half metre rod that is balanced over a pivot using a block of weight 2N and a spring.
Given that the tension in the spring is 9N, determine the weight of the rod.
Answers

Taking moment and equating Clockwise moments = anticlockwise moments
0.6 x 70cm = w x 30
W= 0.6 x 70
30
W = mg = ^{4.2}/_{3}= 1.4N  Since the system is in equilibrium, then (P x A) 15 = w (15+45)
2.0 x10^{5} x 4x 10^{4} x 15 = 60w
w= 8 x 15 x 10
60
Weight, w= 20N  Solution
5kg = 50N
50x = 15 x (3 − x)
50 x = 45 − 15x
50x=4515x
65x= 45
x=^{45}/_{65} = 0.692m  For a system in equilibrium. The sum of clockwise moments about the same point must be equal to the sum of anticlockwise moments about the same point.
 • Steering a wheel in a vehicles
• Tightening a nut using spanner
• Peddling a bicycle
• Opening/closing a door
• Closing /opening a water/gas tap  The distance is small hence the moment produced is not enough to open the door. A lot of force will be required.
 It provides them with the necessary support and also makes them stable as they move about.
 Solution
Clockwise moment = Anticlockwise moment
38 x ^{10}/_{100} = w x ^{25}/_{100}
3.8= 0.25W
W= ^{3.8}/_{0.25}= 15.2N 
 Weight = total upward force
= (10 +20) N
= 30N  Let position of c.o.g be x m away from A i.e
Using point A as pivot thus
30x= 20 × 2
30x=40
x = ^{4}/_{3} = 1.33 m
c.o.g is at 1.33m from A or 0.667 m from B
 Weight = total upward force
 Solution
Sum of clockwise moment = sum of anticlockwise moment
8 x 0.2 = (y  0.25)3 + 0.55
1.6 = 3y 0.75 + 0.55
1.6= 3y  0.2
3y= 1.6 + 0.2 = 1.8m
y= 0.6m
y=60cm
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