EdExcel Mechanics 2 Statics of rigid bodies Chapter Assessment 1. Overhead cables for a tramway are supported by uniform, rigid, horizontal beams of...

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B

beam 1000 N

(i)

1m

m

1000 N

Calculate the tension in the wire.

(ii) Find the magnitude and direction of the force on the beam at A.

[5] [6]

2. A uniform ladder of length 8 m and weight 180 N rests against a smooth, vertical wall and stands on a rough, horizontal surface. A woman of weight 720 N stands on the ladder so that her weight acts at a distance x m from its lower end, as shown in the diagram.

20° 8m 720 N

xm

180 N

The system is in equilibrium with the ladder at 20° to the vertical. (i)

Show that the frictional force between the ladder and the horizontal surface is F N, where F 90(1 x) tan 20 . [5]

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EdExcel Mechanics 2 (ii) Deduce that F increases as x increases and hence find the values of the coefficient of friction between the ladder and the surface for which the woman can stand anywhere on the ladder without it slipping.

[5]

3. A simple lift bridge is modelled as a uniform rod AD of length m and weight 5000 N. The rod is freely hinged at B and rests on a small support at C; AB = m and BC = m, as shown in the diagram below. The bridge closed is represented by the rod being horizontal. A (i)

B

C m

m

D

Calculate the forces acting on the bridge due to the hinge at B and support at C. [5]

A lump of concrete of mass M kg is placed at A to ‘counterbalance’ the bridge to make it easier to open. For the bridge to stay firmly closed, the force at C must be 25 N vertically upwards. (ii) Calculate the value of M.

[4]

With the lump of concrete attached, the bridge is held open at 60° to the horizontal by means of a light rope of negligible mass attached to D. The rope pulls upwards at an angle of 10° to the horizontal, as shown in the diagram below. rope 10°

B

D

60°

A (iii) Calculate the tension in the rope.

[6]

4. A uniform beam AB of length 3 m and weight 80 N is freely hinged at A. Initially, the beam is held horizontally in equilibrium by a small, smooth peg at C

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EdExcel Mechanics 2 where the distance AC is m, as shown in the diagram below. 3m B

A C m (i)

Calculate the force on the beam from the peg at C.

[3]

The peg is now moved so that the beam is in equilibrium with AB at 60° to the vertical, as shown in the diagram below. AC is still m.

B C

60° A

m

(ii) Calculate the new force on the beam due to the peg.

[4]

A light string is now attached to the beam at B. The string is perpendicular to the beam. The beam is in equilibrium with a tension of 20 N in the string, as shown in the diagram below. 20 N B C

60° A

m

(iii) Calculate the new force on the beam due to the peg.

[2]

The peg is now removed and the string attached to a point D vertically above A so that angle ABD is 50°, as shown in the diagram below.

50°

B

60° A

3m

(iv) Calculate the new tension in the string. Calculate also the vertical component of the force acting on the hinge at A. [5]

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EdExcel Mechanics 2 Total 50 marks

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EdExcel Mechanics 2 Solutions to Chapter Assessment 1. C wire

3m

Y A

X

34

3

T

m 1m 1m m 1000 N 1500 N 1000 N

5

B

(i) Taking moments about A: T sin 5 1000 4 1500 1000 0

3 9250 34 1850 34 T 3596 3 The tension in the wire is 3596 N (4 ) 5T

(ii) Resolving horizontally: X T cos 0

9250 1850 34 5 3 34 3 Taking moments about B: Y 5 1000 1500 1000 1 0

X

5Y 8250 Y 1650 9250 2 2 Magnitude of force on beam at A 1650 3497 N (4 ) 3 tan

Y 1650 4950 9250 X 9250 3

Y X

The direction of the force at A is ° (3 ) above the horizontal.

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EdExcel Mechanics 2 2. (i)

S

B 20°

8m 720 N

R

xm

180 N

F A Resolving vertically:

R 720 180 0 R 900

Taking moments about B: 720 sin 20(8 x ) 180 sin 20 4 R sin 20 8 F cos 20 8 0

720(8 x ) 720 7200 sin 20 8F cos 20 0 F 900 90 90(8 x ) tan 20 F 90( 9 8 x )tan 20 F 90(1 x )tan 20 (ii) Since all terms in the expression for x are constant except for x, and x is positive, then as x increases F must increase. If the woman stands at the top of the ladder, x = 8. The maximum frictional force required 90 9 tan 20 810 tan 20 F R

F 810 tan 20 R 900

Since all other forces are vertical, the force at the hinge must be vertical.

3. (i)

Y A

m

B

R m

C

D

m 5000 N

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EdExcel Mechanics 2 Taking moments about B:

5000 0 R 1250

Resolving vertically:

Y R 5000 0 Y 5000 1250 3750

The force at the hinge at B is 3750 N vertically upwards. The force at C is 1250 N vertically upwards. (ii)

Y A

Mg

25 N

B

m

C

m

D

m 5000 N

Taking moments about B:

25 Mg 5000 0 Mg 2940

M 200 (iii)

T 10°

D m

Y m B 60° X m 5000 A 200g

Taking moments about B: 200g cos60 5000cos60 T sin 70 0

sin 70 30 T The tension in the rope is N (3 )

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EdExcel Mechanics 2 4. (i) m A

R

m

B

C 80 N

Taking moments about A: 80 R 0

R 48 The force from the peg is 48 N. (ii) S B 60° A

m

m

C

80 N

Taking moments about A: 80 sin 60 S 0 S The force from the peg is N (3 ) (iii) 20 N Z 60° A

m B

m

C

m 80 N

Taking moments about A: 80 sin 60 Z 20 3 0

Z 120 sin 60 60

Z The force from the peg is N (3 )

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EdExcel Mechanics 2 (iv) 70°

Y A

T 50°

B

60°

X

80 N

Taking moments about A: T sin 50 3 80 sin 60 0 120 sin 60 3 sin 50 The new tension is N (3 )

T

Resolving vertically:

Y T cos 70 80 0 Y 80 cos 70

The vertical component of the force at A is N upwards (3 )

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