## Turning Effect of Forces

## Turning Effect of Forces (O Level)

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Question 1 |

A uniform beam of 1 m is being balanced as shown below. What is the moment of the three newton force about the pivot
(point U)?

10 N cm | |

30 N cm | |

90 N cm | |

180 N cm |

Question 1 Explanation:

Moment about a point = Force X Perpendicular distance from force to the point
Moment of 3 N force about point U
=3 N X (60-50) cm = 30 N cm

Question 2 |

A uniform beam of 1 m is being balanced as shown below. What is moment of the 2 N force about the pivot (point U)?

10 N cm | |

80 N cm | |

90 N cm | |

180 N cm |

Question 2 Explanation:

Moment about a point = Force X perpendicular distance from force to the point
Moment of 2 N force about point U
= 2 N X (90-50) cm =80 N cm

Question 3 |

A uniform beam of 1 m is being balanced as shown below. What is moment of force F about point W?

0 N cm | |

40F N cm | |

50F N cm | |

80F N cm |

Question 3 Explanation:

Moment about a point = Force X perpendicular distance from force to the point
Moment of F N force about point W
= F N X (90-10) cm =80F N cm

Question 4 |

A uniform beam of 1 m is being balanced as shown below. What is moment of force F about point T?

0 N cm | |

40F N cm | |

50F N cm | |

80F N cm |

Question 4 Explanation:

Moment about a point = Force X perpendicular distance from force to the point
Moment of F N force about point T
= F N X (10-10) cm =0 N cm

Question 5 |

A uniform beam of 1 m is supported at the 50 cm mark. Given that a weight of 2 N hangs at the 30 cm mark, how far away from
the pivot must another weight of 4 N be hung to balance the beam?

10 cm | |

15 cm | |

20 cm | |

40 cm |

Question 5 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
4 X L = 2 X (50-30)
L = 10 cm

Question 6 |

A uniform beam of 1 m is supported at the 50 cm mark. Given that a weight of 2 N hangs at the 30 cm mark, at which position must another weight of 2 N be hung to balance the beam?

20 cm mark | |

30 cm mark | |

50 cm mark | |

70 cm mark |

Question 6 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
2 X (x - 50) = 2 X (50-30)
x = 70 cm

Question 7 |

A 1 m long uniform beam is supported at the 50 cm mark. Given that a weight of 2 N hangs at the 30 cm mark, at which position must another weight of 1 N be hung to balance the beam?

40 cm | |

60 cm | |

70 cm | |

90 cm |

Question 7 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
1 X (x - 50) = 2 X (50-30)
x = 90 cm

Question 8 |

A uniform beam of 1 m is supported at the 50 cm mark. Given that a weight of 2 N hangs at 10 cm mark, what is the weight W that hangs at 60 cm mark to balance the beam?

0.5 N | |

2 N | |

4 N | |

8 N |

Question 8 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
W X (60 - 50) = 2 X (50-10)
W = 8 N

Question 9 |

A 1 m long uniform beam is supported at the 50 cm mark. Given that a weight W acting at the 10 cm mark is balanced by a 7 N force and a 3 N force acting downwards at 60 cm mark and 80 cm mark respectively, calculate the weight W.

4 N | |

5 N | |

8 N | |

10 N |

Question 9 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
7 X (60 - 50) + 3 X (80-50) = W X (50-10)
W = 4 N

Question 10 |

A 1 m long uniform beam is supported at the 50 cm mark. Given that a 5 N weight acting at the 10 cm mark is balanced by a force F and a 3 N force acting downwards at 60 cm mark and 80 cm mark respectively as shown, calculate the force F.

2 N | |

3 N | |

7 N | |

11 N |

Question 10 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
F X (60 - 50) + 3 X (80-50) = 5 X (50-10)
F = 11 N

Question 11 |

A 1 m long uniform beam is being balanced as shown. Calculate the force F.

2 N | |

3 N | |

7 N | |

11 N |

Question 11 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (50 cm mark),
F X (60 - 50) + 3 X (80-50) = 5 X (50-20) + 1 X (50-0)
F = 11 N

Question 12 |

A 1 m long uniform beam is supported at the 30 cm mark by a pivot. Given that a mass of 2 kg hangs at the 0 cm mark, at which
position must another mass of 1 kg be hung to balance the beam? (Assume mass of beam = 0 kg)

30 cm mark | |

60 cm mark | |

70 cm mark | |

90 cm mark |

Question 12 Explanation:

Clockwise moment = anti-clockwise moment
Take moment about the pivot (30 cm mark),
2 X (30 - 0) = 1 X (x - 30)
x = 90 cm

Question 13 |

A 1 m long uniform beam of mass 1 kg is supported at the 30 cm mark by a pivot. Given that a mass of 2 kg hangs at the 0 cm mark, at which position must another mass of 1 kg be hung to balance the beam?

30 cm mark | |

60 cm mark | |

70 cm mark | |

90 cm mark |

Question 13 Explanation:

Since the beam is uniform and it is 1 m long, the center of gravity is at 50 cm mark
Take moment about the pivot (30 cm mark)
Clockwise moments = anti-clockwise moments
1 X (50--30) + 1 X (x - 30) = 2 X (30 - 0)
x = 70 cm

Question 14 |

A 1 m long uniform beam of mass 1 kg is supported at the 30 cm mark by a pivot. Given that a mass of 2 kg hangs at the 0 cm mark, what is the mass of the object A hanging at the 70 cm mark to balance the beam?

1 kg | |

2 kg | |

3 kg | |

4 kg |

Question 14 Explanation:

Since the beam is uniform and it is 1 m long, the center of gravity is at 50 cm mark
Take moment about the pivot (30 cm mark)
Clockwise moments = anti-clockwise moments
1 X (50--30) + A X (70 - 30) = 2 X (30 - 0)
A = 1 kg

Question 15 |

A 1 m long uniform beam of weight 1 N is supported at the 30 cm mark by a pivot. Given that a 12 N weight hangs at the 0 cm mark and a 2 N weight hangs at 60 cm mark, what is the weight W hanging at the 100 cm mark to balance the beam?

1 N | |

2 N | |

3 N | |

4 N |

Question 15 Explanation:

Since the beam is uniform and it is 1 m long, the center of gravity is at 50 cm mark
Take moment about the pivot (30 cm mark)
Clockwise moments = anti-clockwise moments
2 X (60 - 30) + W X (100 - 30) + 1 X (50 - 30) = 12 X (30 - 0)
W = 4 N

Question 16 |

A 1 m long uniform beam is being balanced as shown below. Calculate the force F.

1.00 N | |

1.25 N | |

2.50 N | |

5.00 N |

Question 16 Explanation:

Take moment about the pivot (50 cm mark).
Clockwise moment = anti-clockwise moment
2 X (90 - 50) = 3 X (60 - 50) + F X (50 - 10)
F = 1.25 N

Question 17 |

A 1 m long uniform beam is being balanced as shown below. Calculate the force F and the force W.

*Force FÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Force W*1.00 N 2.50 N | |

1.25 N 0.25 N | |

2.50 N 1.50 N | |

5.00 N 4.00 N |

Question 17 Explanation:

Take moment about the 10 cm mark as shown (to elminate the unknown F). The force F acts at the 10 cm mark and does not cause any turning effect about that point and therefore will not be considered in the calculation below.
Clockwise moment = anti-clockwise moment
2 X (90 - 10) = 3 X (60 - 10) + W X (50 - 10)
W = 0.25 N
For an equilibrium system,
sum of upward forces = sum of downward forces
3 + W = 2 + F
3 + (0.25) = 2 + F
F =1.25 N

Question 18 |

A 1 m long uniform beam is being balanced as shown below. Calculate the forces F and G.

*Force FÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Force G*3 N 2 N | |

4 N 3 N | |

5 N 4 N | |

6 N 5 N |

Question 18 Explanation:

Take moment about the 30 cm mark as shown (to eliminate the unknown F). The force F acting at the 30 cm mark does not cause any turning effect about that point and therefore will not be considered in the calculation below.
Clockwise moments = anti-clockwise moments
2 X (50 - 30) + 2 X (100 - 30) = 3 X (70 - 30) + G X (30 - 0)
G = 2 N
For an equilibrium system,
sum of upward forces = sum of downward forces
F + 3 = G + 2 + 2
F + 3 = (2) + 2 + 2
F = 3 N

Question 19 |

3 N | |

4 N | |

5 N | |

6 N |

Question 19 Explanation:

Take moment about the 30 cm mark as shown (to eliminate the unknown F). The force F, force M and 2 N force acting at the 30 cm mark do not cause any turning effect about that point. Therefore, they will not be considered in the calculation below.
Clockwise moments = anti-clockwise momens
3 X (100 - 30) = 3 X (70 - 30) + G X (30 - 0)
G = 3 N

Question 20 |

1 N | |

2 N | |

3 N | |

4 N |

Question 20 Explanation:

Take moment about the 30 cm mark as shown to eliminate the unknown forces. The forces F, M, K, H and N acting at the 30 cm mark do not cause any turning effect about that point and therefore will not be considered in the calculation.
Clockwise moments = anti-clockwise moments
3 X (80 - 30) = G X (60 - 30) + 4 X (30 - 0)
G = 1 N

Question 21 |

A 12 long uniform bridge of negligible weight is being supported at the 2 ends. A lorry weighing 30000 N is parked at the 4 m mark as shown. What are the forces acting on the bridge by support A and support B?
Support AÂ Â Â Â Â Â Â Â Â Â Â Â Â Support B

10000 N 20000 N | |

15000 N 15000 N | |

20000 N 10000 N | |

15000 N 10000 N |

Question 21 Explanation:

Take moment about the 0 m mark as shown (to treat the 0 m mark as a fix point to eliminate the unknown A). The force A acting at the 0 m mark as a result do not cause any turning effect and therefore will not be considered in the calculation below.
Clockwise moments = anti-clockwise moments
30000 X (4 - 0) = B X (12 - 0)
B = 10000 N
For an equilibrium system,
sum of upward forces = sum of downward forces
A + B = 30000
A + (10000) = 30000
A = 20000 N

Question 22 |

A 12 m long uniform bridge weighing 100000 N is being supported at the 2 ends.Â A lorry weighing 30000 N is parked at the 4 m mark as shown. What are the forces acting on the bridge by support A and support B?
Support AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Support B

70000 N 60000 N | |

60000 N 70000 N | |

65000 N 65000 N | |

100000 N 30000 N |

Question 22 Explanation:

Take moment about the 0 m mark as shown (to treat the 0 m mark as a fix point to eliminate the unknown A). The force A acting at the 0 m mark as a result do not cause any turning effect and therefore will not be considered in the calculation below.
Since the beam is uniform and it is 12 m long, its center of gravity is at 6 m mark.
Clockwise moments = anti-clockwise moments
30 000 X (4 - 0) + 100 000 X (6 - 0) = B X (12 - 0)
B = 60 000 N
For an equilibrium system,
sum of upward forces = sum of downward forces
A + B = 30000 + 100000
A + (60000) = 130 000
A = 70000 N

Question 23 |

A windmill is pushed by four external forces as shown.Â What are the possible set of forces for F and G for the windmill to be stationary?
Set A: F=10 N, G=8 N
Set B: F=8 N, G=10 N
Set C: F=2 N, G=16 N

Set A only | |

Set C only | |

Set A and Set B only | |

Set A, Set B and Set C |

Question 23 Explanation:

There are at least 5 forces acting on the propeller. On top of the 4 forces shown, there must be one or more forces acting at the centre of the propeller caused by the axle.
Take moment at the centre of the propeller to eliminate all unknown forces at the centre,
Clockwise moments = anti-clockwise moments
2G + 2F = 2 x 10 + 2 x 8
G + F = 18
All three sets of forces satisfy G + F = 18.

Question 24 |

A windmill is pushed by four external forces as shown. Calculate Force F required to make the windmill stop moving.

-10 N | |

10 N | |

13 N | |

17 N |

Question 24 Explanation:

There are at least 5 forces acting on the propeller. On top of the 4 forces shown, there must be one or more forces acting at the centre of the propeller caused by the axle.
Take moment at the centre of the propeller to eliminate all unknown forces at the centre,
Clockwise moments = anti-clockwise moments

Question 25 |

A windmill is pushed by four external forces as shown. Calculate Force F required to make the windmill stand still.

2 N | |

4 N | |

6 N | |

16 N |

Question 25 Explanation:

There are at least 5 forces acting on the propeller. On top of the 4 forces shown, there must be one or more forces acting at the centre of the propeller caused by the axle.
Take moment at the centre of the propeller to eliminate all unknown forces at the centre,
Clockwise moments = anti-clockwise moments
3 x 2 + 2 x 3 + 1 x 4 = F x 4
F = 4 N

Question 26 |

3.0 N | |

4.5 N | |

5.0 N | |

6.0 N |

Question 26 Explanation:

Take moment about the 30 cm mark to eliminate the unknown G. The force G and 3 N force acting at the 30 cm mark do not cause any turning effect about the point and therefore will not be considered in the calculation of moments below.
Clockwise moments = anti-clockwise moments
F x (90 - 30) = 6 x (50 - 30)
F = 2 N
For an equilibrium system, the resultant force acting on the beam must be zero.
Vertical component of G = 6 - 2 = 4 N
Horizontal component of G = 3 N
G x G = (4 x 4) + (3 x 3)
G = 5 N

Question 27 |

A uniform beam of 2 m is fixed to a wall and loaded by the forces shown below. Given that the beam is at equilibrium, calculate force F.

2 N | |

4 N | |

7 N | |

8 N |

Question 27 Explanation:

Take moment about the 0 m mark as shown (to treat the 0 m mark as a fix point to eliminate the unknown forces acting by the wall on the beam). The unknown forces and the 3 N forces acting on the 0 m mark as a result do not cause any turning effect and therefore will not be considered in the calculation of moment below,
Clockwise moment = anti-clockwise moment
F x (2 - 0) = 4 x (1 -0)
F = 2 N

Question 28 |

A uniform beam of 2 m is fixed to a wall and loaded by the forces shown below. Given that the beam is at equilibrium, calculate the resulting force acting at the 0 m mark.

2.0 N | |

3.0 N | |

3.6 N | |

5.0 N |

Question 28 Explanation:

Let the unknown forces acting at the 0 m mark of the beam by the wall be x & y.
Take moment about the 0 m mark to eliminate the unknown forces x & y of the wall on the beam. The unknown forces x, y and the 3 N forces acting at the 0 m mark do not cause any turning effect about that point. therefore they will not be considered in the calculation of moments below.
Clockwise moment = anti-clockwise moment
F x (2 -0) = 4 x (1 - 0)
F = 2 N
For an equilibrium system, the resultant force acting on the beam must be zero.

Question 29 |

A 1 m long uniform beam of 2 kg mass is being lifted vertically up by a force F at the 100 cm mark. What is the minimum force to do so?

1 N | |

2 N | |

10 N | |

20 N |

Question 29 Explanation:

Let the unknown forces acting at the 0 cm mark of the beam by the floor be R as shown in the diagram
Take moment about the 0 cm mark to eliminate the unknown force R. The unknown force R acting at the 0 m mark does not cause any turning effect about that point and therefore will not be considered in the calculation of moments below.
Weight of beam = mg = 2 x 10 = 20 N
clockwise moment = anti-clockwise moment
20 x (50 - 0) = F x (100 - 0)
F = 10 N

Question 30 |

A 1 m long uniform beam of 6 N weight is being lifted up vertically by a force F at the 75 cm mark. What is the minimum force to do so?

2 N | |

4 N | |

6 N | |

8 N |

Question 30 Explanation:

Let the unknown force acting at the 0 cm mark of the beam by the floor be R as shown in the diagram.
take moment about the 0 cm mark to eliminate the unknown force R. The unknown force R acting at the 0 cm mark does not cause any turning effect about that point and therefore will not be considered in the calculation of moments below.
Weight of beam = 6 N
Clockwise moment = anti-clockwise moment
6 x (50 - 0) = F x (75 -0)
F = 4 N

Question 31 |

Which of the following containers shown below is more stable given that the density of the liquid is much higher than the density of the container?

A | |

B | |

C | |

D |

Question 31 Explanation:

The centre of gravity of choice A is at the centre of the container.
When small amount of liquid is poured into the container, the weight of the liquid concentrates at the lower end of the container. This will lower the centre of gravity of the total object more stable.
As more liquid is being poured into the container, the centre of gravity of the liquid goes higher and therefore the total object will get more unstable.
When the liquid is filled to the brim, in choice D, it will be as less stable as choice A.

Question 32 |

The diagrams below showed the position of the centre of gravity of each standing lamina. Given that the laminas have the same weight, which of the laminas shown below is the most stable?

A | |

B | |

C | |

D |

Question 32 Explanation:

The degree of stability depends on how much angle to tilt before the centre of gravity is not being supported by the base. the more stable an object is, the greater the angle needed to be tilted before the object falls to another position.

Question 33 |

A man of mass 50 kg starts walking from pivot B towards the right edge of a 100 N uniform beam (4 m) as shown below. Given that the beam is not fixed to the two pivots, how far can the man walk before the beam topple?

0.1 m | |

0.2 m | |

0.4 m | |

0.5 m |

Question 33 Explanation:

The beam will topple when the man reaches a point where the beam lifts off from pivot A. At this moment, pivot A does not apply any force on the beam.
Take moment about the point acted by pivot B.
Weight of beam = 100 N
Weight of man = 500 N
Clockwise moment = anti-clockwise moment
500 x (x) = 100 x (1)
x = 0.2 m

Question 34 |

A horizontal force

*F*is applied to a cylinder weighing 100 N as shown below. Given that the radius of the cylinder is 0.5 m, what is the smallest value of*F*to cause the cylinder to roll up the step?25 N | |

50 N | |

75 N | |

100 N |

Question 34 Explanation:

The cylinder will start to row roll up only when the reaction force R1 of the floor on the cylinder is zero (just before the cylinder lift up from the floor, the floor is not supporting the cylinder).
The length OB = radius - 0.1
= 0.5 - 0.1
= 0.4 m
The length OA = radius
= 0.5 m
The length OC is calculated by
(OA x OA) = (OC x OC) + (OB x OB)
(0.5 x 0.5) = (OC x OC) + (0.3 X 0.3)
OC = 0.4 m
Take moment about point A to eliminate R2,
Clockwise moment = anti-clockwise moment
100 x (0.3) = F x (0.4)
F = 75 N

Question 35 |

A kite hangs freely on the branch of a tree as shown. Which point is possibly the centre of gravity of the Kite?

A | |

B | |

C | |

D |

Question 35 Explanation:

For a hanging object at equilibrium, the centre of gravity must be vertically below the pivot point. When this happens, the perpendicular distance between the force and the pivot point is zero and therefore there is no moment created to set the object in clockwise or anti-clockwise motion.

Question 36 |

A box is balanced at the edge of a table as shown. Which point is possibly the centre of gravity of the box?

A | |

B | |

C | |

D |

Question 36 Explanation:

For an object resting on an area, the centre of gravity of the object must must be within the contact area between object and the support. This will allow the reaction force of the supporting surface to balance the weight of the object directly.

Question 37 |

A 10 kg box, Â a 20 kg box and a 30 kg box are stacked up. Which of the following combinations will be most unstable?

A | |

B | |

C | |

D |

Question 37 Explanation:

the most unstable case is when the centre of gravity is at the highest position.

Question 38 |

The diagram below shows a balancing toy pivoting on a triangular stand. When the toy is tilted slightly, it swings back to its original position. Where is the centre of gravity of the toy?

A | |

B | |

C | |

D |

Question 38 Explanation:

Whenever an object balanced with its centre of gravity at a lower position than the pivoting point, is being slightly displaced, it will always rotate back to its original position. This kind of equilibrium is known as stable equilibrium.

Question 39 |

The diagram below shows a balancing toy pivoting on a triangular stand. When the toy is tilted slightly, it stays at the new position. Where is the centre of gravity of the toy?

A | |

B | |

C | |

D |

Question 39 Explanation:

Whenever an object balanced with its centre of gravity at a lower position than the pivoting point, is being slightly displaced, it will always stay at its newly displayed position. This kind of equilibrium is known as neutral equilibrium.

Question 40 |

A uniform metre rod of weight 10 N leans against a smooth glass door in an equilibrium position as shown. What is the reaction force of the glass door on the rod?

4.3 N | |

5.0 N | |

8.7 N | |

10.1 N |

Question 40 Explanation:

Take moment about point A to eliminate F1 and F2.
Clockwise moment = anti-clockwise moment
10 x (0.5 x cos 30) = R x (1 x sin 30)
R = 8.66 N

Question 41 |

Two downward forces, F and 30 N, act on a light uniform plank of length L at the two ends as shown in the diagram. What is the magnitude of the upward force acting on the plank at the pivot?

10 N | |

30 N | |

40 N | |

120 N |

Question 41 Explanation:

The upward force acting on the plank is the force of the pivot R.
Take moment about point A to eliminate F.
Clockwise moment = anti-clockwise moment
30 x L = R x (0.75 L)
R = 40 N

Question 42 |

A 100 N uniform rod of length 6L is supported by two forces

*F*and_{1 }*F*as shown in the diagram. What is the ratioÂ_{2 }*F*Â_{1 }:*F*?_{2}1 : 2 | |

2 : 1 | |

1 : 3 | |

3 : 1 |

Question 42 Explanation:

Take moment about point A to eliminate 100 N force.
Clockwise moment = anti-clockwise moment
F1 x (2L) = F2 x (L)
F1 / F2 = L / 2L
F1 / F2 = 1 / 2

Question 43 |

A man uses a metal rod to raise a large stone as shown below. Which force would raise the large stone most easily?

A | |

B | |

C | |

D |

Question 43 Explanation:

To create the maximum moment to raise the large stone, the perpendicular distance from the force to the pivot point must be long. The longest perpendicular distance is from force C to the pivot point.

Question 44 |

A uniform rod is held vertically as shown. Where should the rod be held such that when it is slightly displaced, it drops to another new position?

position 2 only | |

position 2 only | |

position 1 and position 3 only | |

position 1, position 2 and position 3 |

Question 44 Explanation:

Whenever an object balanced with its centre of gravity at a higher position than the pivoting point, is being slightly displaces, it will always drop to a new position with the centre of gravity at a lower position. This kind of equilibrium is known as unstable equilibrium. Only position 3 is lower than the centre of gravity.

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