EdExcel Mechanics 2

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Nov 26, 2013 (3 years and 8 months ago)

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EdExcel M
echanics 2


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Statics of rigid bodies


Chapter Assessment


1.

Overhead cables for a tramway are supported by uniform, rigid, horizontal beams
of weight 1500 N and length 5 m. Each beam, AB, is freely pivoted at one end A
and supports two cables which may be modelled by ver
tical loads, each of

1000 N, one 1.5 m from A and the other at 1 m from B. The beam is supported by
a light wire, attached at one end to the beam at B and at the other to the point C
which is 3 m vertically above A, as shown in the diagram below.


(i)

Calculate the tension in the wire.






[5]


(ii)

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


[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.


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


(i)

Show that the f
rictional force between the ladder and the horizontal surface is
F

N, where





90(1 ) tan20
  
F x
.





[5]


720 N

180 N

x

m

8 m

20°

A

B

C

3 m

beam

wire

1.5 m

2.5 m

1 m

1000 N

1000 N


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(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 whi
ch 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 4.2 m and weight
5000 N. The rod is freely hinged at B and rests on a small support at C;

AB = 1.5 m and BC = 2.4 m,
as shown in the diagram below. The bridge closed is
represented by the rod being horizontal.





(i)

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 ‘counterbalan
ce’ 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 m
eans 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.















(iii)

Calculate the tension in the rope.






[6]



4.

A uniform beam AB of length 3 m and weight 80 N i
s freely hinged at A.

Initially, the beam is held horizontally in equilibrium by a small, smooth peg at C
A

B

C

D

1.5 m

2.4 m

A

B

D

60°

10°

rope


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where the distance AC is 2.5 m, as shown in the diagram below.


(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 2.5 m.






(ii)

Calculate the new force on the beam due to the peg.



[4]


A light string is now att
ached 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.






(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.






(iv)

Calculate the new tension in the string. Calculate also the vertical component
of the forc
e acting on the hinge at A.





[5]


A

B

60°

3 m

50°

A

B

C

60°

2.5 m

20 N

A

C

3 m

2.5 m

B

A

B

C

60°

2.5 m


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Total 50 marks




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Solutions to
Chapter Assessment


1.




(i)

Taking moments about A:




sin 5 1000 4 1500 2.5 1000 1.5 0
3
5 9250
34
1850 34
3596
3
T
T
T

       
 
 



The tension in the wire is
3596

N (4 s.f.)


(ii)

Resolving horizontally:





cos 0
9250
1850 34 5
3 34 3
X T
X

 
  



Taking moments about B:





5 1000 3.5 1500 2.5 1000 1 0
5 8250
1650
Y
Y
Y
       





Magnitude of force on beam at A
2
2
9250
1650 3497
3
 
  
 
 

N (4 s.f.)




9250
3
4950
1650
tan
9250
28.2
Y
X


  
 







The direction of the force at A is 28.2° (3 s.f.) above t
he horizontal.




A

B

C

3 m

wire

1.5 m

1
.5 m

1 m

T

1000 N

1000 N

X

Y





5

3

34

1 m

1500 N



Y

X


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2.

(i)






Resolving vertically:


720 180 0
900
R
R
  




Taking moments about B:








720sin20 (8 ) 180sin20 4 sin20 8 cos20 8 0
720(8 ) 720 7200 sin20 8 cos20 0
900 90 90(8 ) tan20
90( 9 8 )tan20
90(1 )tan20
x R F
x F
F x
F x
F x
        
      
    
   
  


(ii)

Since all terms in the expression for x are constant except for x, and x is



positiv
e, 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 9tan20 810tan20
    



810tan20
900
0.328
F R
F
R





 




3.

(i)


720 N

180 N

x

m

8 m

20°

S

F

R

A

B

A

B

C

D

1.5 m

2.4 m

2
.
1

m

5000 N

Y

R

Since all other forces

are
vertical, the force at the
hinge must be vertical.


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Taking moments about B:

2.4 0.6 5000 0
1250
R
R
  






Resolving vertically:


5000 0
5000 1250 3750
Y R
Y
  
  




The force at the hinge at B is 3750 N vertically upwards.



The force at C is 1250 N vertically upwards.


(ii)






Taking moments about B:

25 2.4 1.5 5000 0.6 0
1.5 2940
200
Mg
Mg
M
    




(iii)





Taking moments about B:




200 cos60 1.5 5000cos60 0.6 sin 70 2.7 0
2.7 sin 70 30
11.8
g T
T
T
     
 




The tension in the rope is
11.8

N (3 s.f.)


A

B

C

D

1.5 m

2.4 m

2
.
1

m

5000 N

Y

25 N

Mg

A

B

D

60°

10°

200
g

50
00

T

Y

X

1.5 m

0.6 m

2.1 m


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4.

(i)





Taking moments about A:

80 1.5 2.5 0
48
R
R
  




The force from the peg is 48 N.


(ii)







Taking moments about A:

80sin60 1.5 2.5 0
41.6
S
S
   




The force from the peg is 41.6 N (3 s.f.)


(iii)





Taking moments about A:

80sin60 1.5 2.5 20 3 0
2.5 120sin60 60
17.6
Z
Z
Z
     
  




The force from the peg is 17.6 N (3 s.f.)


A

B

C

60°

2.5 m

20 N

Z

0
.5 m

80 N

1
.5 m

A

C

1.5

m

2.5 m

B

80 N

R

A

B

C

60°

2.5 m

1
.5 m

80 N

S


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(iv)






Taking moments about A:

sin50 3 80sin60 1.5 0
120sin60
45.2
3sin50
T
T
   

 




The new tension is 45.2 N (3 s.f.)




Resolving vertically:

cos 70 80 0
80 45.22cos 70 64.5
Y T
Y
   
   



The vertical component
of the force at A is 64.5 N

upwards

(3 s.f.)








A

B

60°

80 N

50°

T

X

Y

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