O lvl Phy: Moments

The figure below shows a uniform rod AB, 100.0 cm long, with a mass m of 225 g hung from the end A. The rod is balanced on a knife edge place 20.0 cm from A.

(i) Explain carefully why the rod, with the mass attached rests in a horizontal position.
(ii) Calculate the mass of the rod
(iii) If m is increased to 260 g, where should the knife-edge be placed such that the rod remains balanced.

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Answer:

(i) The total anti-clockwise moments about the knife-edge is equal to the total clockwise moments about the knife-edge.

(ii) Let the mass of rod = y
Note: The centre of gravity of the rod is 50 cm - 20 cm = 30 cm from the knife-edge
The mass m is 20 cm from the knife-edge

Using the principle of moments,
total anti-clockwise moments about the knife-edge = total clockwise moments about the same knife-edge
0.225 * 10 * 0.20 = y * 10 * 0.3
y = 0.15 kg

Hence, mass of rod is 150 g

(iii) Let the new distance of the knife-edge from A = d
Hence,
0.260 * 10 * d = 0.15 * 10 * (0.5 - d)
2.6d = 0.75 - 1.5d
4.1d = 0.75
d = 0.183 m

Knife-edge should be placed 18.3 cm away from A.

O lvl Phy: Moments

The mass of the uniform beam (gray coloured) is 15 kg. Calculate the value of distance x m, that a 20 kg mass must be placed in order for the tension in string A to be twice the tension in string B.

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Answer:

Using principle of moments, and taking moments about A,
sum of clockwise moments = sum of anti-clockwise moments

200 * x + 150 * 4 = F * 8
200x + 600 = 8F --------- (1)

Total upward force = total downard force

2F + F = 200 + 150
3F = 350
F = 116.67 N -------- (2)

Thus, substituting (2) into (1)
200x + 600 = 8 (116.67)
x = 1.67 m

O lvl Phy: Moments

The diagram shows a wrench turning a nut.

(a) State the principle of moments for a body in equilibrium

(b) Explain why it is easier to turn the nut with a wrench than with your fingers.

(c) Given that the diameter of the bolt is 1.2 cm and the maximum friction between the nut and the bolt is 48.0 N, find the minimum force the hand needs to exert on the wrench. The maximum distance of the centre of the bolt to the end of the wrench is 24.0 cm. Draw a clearly labelled diagram to show the forces involved.

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Answer:

(a) For a body in equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anti-clockwise moments about the same pivot.

(b) It is easier to turn the nut with a wrench than with your fingers because the distance from the centre of the bolt to where the force is exerted is larger when using the wrench than when using the fingers. Hence, for the wrench, a smaller force is required to create the same turning effect as compared to using the fingers.

(c) Using the Principle of Moments,
and taking moments about the centre of the bolt,
F * 24.0 cm = 48.0 N * 0.6 cm (radius of bolt)
Resolving,
F = 1.2 N

O lvl Phy: Moments

A non-uniform plank weighing 300N is set up as shown. The spring balance reads 160 N when the small boy stands at A, and 760 N when the small boy stands at B.

a) State what is meant by moment of a force.

b) Calculate the weight of the boy.

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Answer:

a) The moment of a force is the turning effect of the force. It is calculated by the multiplication of the force and the perpendicular distance from the pivot.

b)

Let the weight of the boy be W.
Let the distance from the pivot to the c.g. of the plank be x.

When the boy is at A, taking pivot about pivot point (so that R can be ignored)

Clockwise moments = 300x

Anti-clockwise moments = 160 * 5 + 2W

At equilibrium, clockwise moments = anti-clockwise moments

300x = 800 + 2W --------------------------- (1)

Note: R is ignored because the perpendicular distance between R and the pivot is 0, i.e. moment of R about pivot is 0. This is a common thing to do in physics to eliminate unknown forces, so as to cut down on the number of unknowns.

Clockwise moments = 300x + 8W

Anti-clockwise moments = 760 * 5

At equilibrium, clockwise moments = anti-clockwise moments

300x + 8W = 3800 --------------------------- (2)

Sub (1) into (2):
800 + 2W + 8W = 3800
10W = 3000
W = 300

Hence, the weight of the boy is 300 N