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 m2 and the pressure of the steam in the boiler is 2.0 x 105 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 x105 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=45-15x
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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