1. A seesaw in a playground consists of a beam AB of length 4 m which is supported by a smooth pivot at its centre C. Jill has mass 25 kg and sits on the end A. David has mass 40 kg and sits at a distance x metres from C, as shown in the figure. The beam is initially modelled as a uniform rod. Using this model,
a. Find the value of x for which the seesaw can rest in equilibrium in a horizontal position (3) b. State what is implied by the modelling assumption that the beam is uniform. (1) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments (Equilibrium) Chapter Reference: Mechanics 2, Chapter 4
5 minutes
Solutions
1a. Moments about C M1 25g x 2 = 40g x x M1 x = 1.25m M1
1b.
Weight/mass acts at mid-point; or weight/mass evenly distributed M1
1. A beam AB has length 15 m. The beam rests horizontally in equilibrium on two smooth supports at the points P and Q, where AP = 2 m and QB = 3 m. When a child of mass 50 kg stands on the beam at A, the beam remains in equilibrium and is on the point of tilting about P. When the same child of mass 50 kg stands on the beam at B, the beam remains in equilibrium and is on the point of tilting about Q. The child is modelled as a particle and the beam is modelled as a non-uniform rod. a. Find the mass of the beam. b. Find the distance of the centre of mass of the beam from A. (8) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments (Horizontal Beans) Chapter Reference: Mechanics 2, Chapter 4
7 minutes
Solutions
1a.
M1
Taking moments abouts P M1 50g x 2 = mg x (x β 2) M1 Taking moments about Q M1 50g x 3 = mg x (12 β x) M1 Solving by elimination M1 m = 25 kg M1 x = 6 M1
1. A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC = DB = d, as shown in the figure. The centre of mass of the rod is at the point G. A particle of mass 5
2
m is placed on the rod at B and the rod is on the point of tipping about D. a. Show that GD = 5
2ππ (4)
The particle is moved from B to the mid-point of the rod and the rod remains in equilibrium. b. Find the magnitude of the normal reaction between the support at D and the rod. (4) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments and Tilting Beams Chapter Reference: Mechanics 2, Chapter 4
8 minutes
Solutions
1a.
M1
Taking moments about D M1 mg x GD = 5
2ππππ Γ ππ M1
GD = 52ππ M1
1b.
M1
Taking moments about C M1 mg x ππ
2+ 5
2ππππ Γ 3
2ππ = ππ Γ 3ππ M1
Y = 1712ππππ M1
1. A uniform rod AB has mass 4 kg and length 1.4 m. The end A is resting on rough horizontal ground. A light string BC has one end attached to B and the other end attached to a fixed point C. The string is perpendicular to the rod and lies in the same vertical plane as the rod. The rod is in equilibrium, inclined at 20Β° to the ground, as shown in the figure. a. Find the tension in the string (3) Given that the rod is about to slip. b. Find the coefficient of friction between the road and the ground. (7) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments of a Non-Perpendicular Force Chapter Reference: Mechanics 2, Chapter 4
8 minutes
Solutions
1a. Taking moments about A M1 4g x 0.7 x cos 20 = 1.4T M1 T = 18.4N M1
1b.
Resolving vertically M1 R + T cos 20 = 4g R = 4g β T cos 20 M1
Resolving horizontally M1 F = T sin 20 M1 F = ππR F = T sin 20 M1
T sin 20 = ππ(4ππ β ππ ππππππ20) M1 ππ = ππ sin20
4ππβππ cos20= 0.29 M1
1. A ladder, of length 5 m and mass 18 kg, has one end A resting on rough horizontal ground and its other end B resting against a smooth vertical wall. The ladder lies in a vertical plane perpendicular to the wall and makes an angle πΌπΌ with the horizontal ground, where tan Ξ± = 4
3 , as shown in the figure. The coefficient of friction between
the ladder and the ground is Β΅. A woman of mass 60 kg stands on the ladder at the point C, where AC = 3 m. The ladder is on the point of slipping. The ladder is modelled as a uniform rod and the woman as a particle. Find the value of Β΅ (8) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments (Ladder Problem) Chapter Reference: Mechanics 2, Chapter 4
12 minutes
Solutions
1. πΉπΉ = ππππ M1 Resolving vertically 18g + 60g = ππ ππ = 78g
M1
Resolving horizontally π π = πΉπΉ = ππN M1
2.5 x 18g cos πΌπΌ + 3 x 60g cos πΌπΌ = 5F sin πΌπΌ 18g x 2.5 cos πΌπΌ + 60ππ x 3 cos πΌπΌ = R x 5 sin πΌπΌ 12
cosπΌπΌ x 18g + 3 sin πΌπΌ πΉπΉ + 2 sinπΌπΌ π π = 3 cos πΌπΌ ππ M1
5 cos πΌπΌ N = 5 sin πΌπΌ πΉπΉ + 2.5 cos πΌπΌ x 18g + 2 cos πΌπΌ x 60 60g x 1
2cos πΌπΌ + 2.5N cos πΌπΌ = 2.5R sin πΌπΌ + 2.5F sin πΌπΌ M1
45 x 35ππ + 180 Γ 3
5ππ = 4R
R = 1354
g M1
78gππ = 1354
g M1
ππ = 1354Γ78
= 135312
= 0.43
M1
1. A steel girder AB has weight 210 N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end A. The other cable is attached to the point C on the girder, where AC = 90 cm, as shown in the figure. The girder is modelled as a uniform rod, and the cables as light inextensible strings. A small load of weight W newtons is attached to the girder at B. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position. The tension in the cable at C is now three times the tension in the cable at A. Find the value of W. (6) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Moments (Solving Weight) Chapter Reference: Mechanics 2, Chapter 4
6 minutes
Solutions
1a.
Resolving vertically M1 4S = 210 + W M1 Taking moment about B (o.e) M1 S x 120 + 3S x 30 = 210 x 60 M1 Solving simultaneously M1 (S = 60) W = 30 M1