# Equilibrium of Bodies

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```					Chapter 2

2.1   (i)

y
T(y)

L
y

M
T(y+y)+     yg
L

(ii) Since the element of length y is in equilibrium, we have
M
T ( y)  T ( y  y)         yg
L
Using Taylor series expansion for T(y+y), which gives
dT      1 d 2T
T ( y  y)  T ( y)       y       2
(y) 2  
dy      2 dy
And taking limit y  0 leads to the differential equation for T(y). The equation
is

dT   M
 g
dy   l
Solution of this equation is
M
T ( y)       gy  C
L
where C is the integration constant. This is determined by using the fact that at the
lose end (y=L) of the rope, the tension is zero. This gives
M ( L  y)
C  Mg and T ( y)               g
L

  
2.2      Torque   r  F
    ˆ j
It is given that r  2i  ˆ and the force has magnitude 50N and acts in the direction
ˆ
of vector 3i  2 ˆ . Thus the force is 50 times the unit vector in the direction of the
j
given vector. This gives
        ˆ
 3i  2 ˆ 
j
 13 
F  50          
          


ˆ j
With this the torque is 2i  ˆ 50 ˆ

3i  2 ˆ 
j   
50 ˆ
k
13             13
2.3      (a)

100N

100N

(b) The centre of the rod is at (3, 2)
    ˆ
The right end is at position r  8i  2 ˆ and the force at this end is
j

        3         
F  100      ˆ 1 ˆ
 2 i  2 j
           
     ˆ
The left end is at position r  2i  2 ˆ and the force at this end is
j
        3          
F  100   ˆ 1
i      ˆ
j
 2    2
            
Torque with respect to the origin =

8iˆ  2 ˆj 100


3ˆ 1
i


ˆ    2i  2 ˆ  100 3 i  1 ˆ   10i  100 3 i  1 ˆ 
j       ˆ     
j

 2
ˆ     j

ˆ

 2
ˆ     j

 2    2                                2                      2 
 500k  ˆ

Torque with respect to the centre of the rod =
 3                        3                      3         
8i  100
ˆ          ˆ 1 ˆ         
ˆ          ˆ 1 ˆ        ˆ         ˆ 1 ˆ
 2 i  2 j    2i  100 2 i  2 j   10i  100 2 i  2 j 
                                                           
 500k ˆ

Torque with respect to the left end of the rod =
 3        
10 i  100 
ˆ           ˆ 1 ˆ
 2 i  2 j
          
 500 k ˆ

Torque with respect two the right end of the rod =
 3          
 10 i  100 
ˆ            ˆ 1     ˆ
 2 i2     j
            
 500 k ˆ

(c) Torques about all the points are equal because the net force on the rod is zero.
2.4 (a)

TA                                          TB

3m
A                                                    B

1m                                0.5m
30N

70N                    120N

(b)   F   y    0 gives

TA  TB  220                            (i)
Net torque about A is zero, which gives
3TB  1  70  1.5  30  2.5  120
(ii)
 415

Equation (ii) gives TB  138 N

This substituted in equation (i) gives TA  82 N

2.5       Component of force in the plane perpendicular to the axis is F cos at a distance of
R from the axis.
Therefore the torque about the axis is RF cos

2.6 Free-body diagram of the block
N1

N2

W

N1, N2 and W are three forces in a plane. Thus they must pass through one common
point for equilibrium. So the equilibrium conditions are only the force conditions.

F  horizontal    0 gives

N1 sin   N 2 cos

F  vertical    0 gives

N1 cos  N1 sin   W
Solution of these two equations is
N1  W cos   and N 2  W sin 
2.7 Free-body diagram of the plank

N                      2m

Ry
1m                                                      100N

0.2m

Rx

Free-body diagram of the block

N

F

Nground

If the rod makes angle  with the horizontal then
sin   0.2 and cos  0.98
(a) To get the horizontal force F, we first calculate the normal force N on the rod. To do so,
we calculate the total torque about the hinged end of the plank  1  N  3  100 cos  and
equate it to zero. This gives
N  3  100 cos
 3  98
 294N

Now we balance the horizontal forces    F     horizontal    0 on the block to get

F  294 sin 
 0.2  294
 59 N
(b) Force balance on the plank

F    horizontal    0  Rx  N sin   59N

F      vertical    0  Ry  N cos  100
This gives
R y  100  288  188 N

Minus sign in front implies that the direction is opposite to that shown in the free-body
diagram above.

2.8    Free-body diagram of the rod

N2

N1

F
W

Balancing the vertical forces gives N1 = W = 50N
Balancing the horizontal forces gives N2 = F
Balancing the torque about the centre of gravity gives
F  8  50  0.5
25
F         8.8 N
8
2.9    Free body diagram of the painting
T

N1           N2

N1+N2

Fx

W                                  W

Force balance equations give
Fx  T and N1  N 2  W

N1 and N2 are equal because the component of torque perpendicular to the wall must vanish.
This gives
N1=N2=25N
Balancing the component of torque parallel to the wall taken about the lower end of the
painting gives
20 3  T  10  50
giving
25
T         14 .4 N
3

2.10     We first calculate the forces at the ends of the rod. These forces are applied by the
supports. After finding the forces on the rod, we then calculate the forces and the
torques applied by the wall on the supports.

Free body diagram of the rod
N1                                                           N2
140cm

60cm

35N

Free body diagrams of the left and the right supports

F1                                        F2

1                                         2
5cm                                        5cm
N1                                       N2

The forces on the rod satisfy   Fy    0 which gives

N1  N 2  35

Taking torque about the left end and using       0 gives
140  N 2  60  35  N 2  15 N
This gives
N1  20 N
Now balancing vertical forces and the torque on the supports gives
For the left support F1=20N and  1  0.05  20  1Nm

For the right support F2=15N and  2  0.05  15  0.75 Nm

2.11
Ry                60cm

Rx

40cm

N

40N

To find the force applied by the plastic block, we balance torque about the upper left corner.
40  N  30  40  N  30N
Balancing the vertical forces gives Ry = 40N
Balancing the horizontal forces gives Rx = 30N
Negative sign means that the direction of Rx is opposite to that assumed in the free-body diagram
above.
Free-body diagram of the pole
30N

40cm
40N

30N

N



Balancing the vertical forces on the pole gives N = 40N
There is no net horizontal force and the two horizontal forces give a couple = 300.4 = 12Nm
Balancing the torques on the pole about the ground gives  = 300.4 = 12Nm

2.12       Free-body diagram of the table

90cm

ˆ
j
Rx                                        Nx

Ny
Ry
iˆ                                  20N

To find Ny, we balance the torque on the table about its left hand edge to get
90  Ny  45  20  Ny  10N
By balancing the vertical forces, we get Ry  10N . The negative sign tells us that the force is
direction opposite to that shown above.
Free-body diagram of one of the rods

Ry
2

30
Rx
2

Sx

Sy

Free body diagram of the entire system

90cm

30                                  Nx

Ny

20N
2Sy              2Sx

To get Nx, we balance the component of the torque coming out of the paper on the entire system
about the lower hinge. This gives
30 3  Nx  45  20        Nx  10 3N
The negative sign again tells us that the direction of the force is opposite to that shown.
Balancing the horizontal component of the force on the table then gives Rx  10 3N
Rx ˆ Ry ˆ
Note: The net force on each rod on its upper end is              i          ˆ
j  5 3i  5 ˆ which is along
j
2     2
the rod as it must be for the equilibrium of a rod held at its ends.
Balancing the horizontal and vertical components of forces on each rod gives
Rx
Sx        5 3N and Sy  5N
2
Thus the net force on each rod is 10N compressive.

2.13 Free body diagrams of the two side portions and the portion AC over the pulley:

N
TA                                                  TC

TA
TC
RM
g
L

F
L2 M
g
L
L1 M
g
L

Tension TA and TC at both ends of the portion over the pulley is the same because the torque
L1
about the centre must vanish. This gives TA  TC             Mg
L
Free body diagrams of the portion AB and BC
Neffy                                        Neffy

Neffx                                                                       Neffx
TB                    TB

RM                                         RM
g                                          g
2L                                           2L

TA                                                                     TC

Notice that Torque of the normal reaction about the centre of the cylinder vanishes because for
each small portion of the rope over the cylinder, the normal reaction is radial. Thus T A (or TC)
and TB cannot be equal because they together provide a torque to balance the torque due to the
weight of the rope. Balancing the torque about the centre on AB gives
R         RM                               R  M
R  TA                   g  R  TB     TB   L1       g
2       2L                               2 2 L
Thus if the net force by the cylinder on the rope is Neff at an angle  from the horizontal then by
force balance

      R  M                                 R  M
N eff sin    L1      g             N eff cos   L1       g
       2  L                                2 2 L
Note that Neff acts at a point different from the centre of BC because on different infinitesimal
portions it is different.

2.14 The support does not apply any torque about the x-axis. All other components and torques
are balanced by the support.

2.15 When forces are applied at two points of the rod, force balance demands that the force be
equal and opposite. However two such forces acting at two different points will give rise to
a couple moment. The couple moment is zero only if the forces point along the rod (see
figure below)
Couple moment non-zero                            Couple moment zero

2.16 Let cables OA and OC make angle 1 and OB and OD angle 2 with the vertical. Then
balancing the vertical forces gives
2T (sin 1  sin  2 )  45000
The sine of the angles is easily calculated to be
1                  1            2
sin 1         sin  2               
2               11 4          5
This gives T=14050N


2.17 The torque direction is given by the direction of cross product n  F , which is
ˆ
ˆ                                                                        ˆ
perpendicular to n . This implies there is no component of the torque in the direction of n .

ˆ             ˆ       ˆ
2.18 The net force on the plate is 50 i  50 ˆ  70 i  120 i  50 ˆ
j                     j
Therefore the force that must be applied to the plate to keep it in equilibrium is

ˆ
F  120 i  50 ˆ . Since there are only three forces acting on the body, they must all pass
j
through the same point so that their net torque is zero. This is shown in figure below.
A                                 B

O

D                                   C

                                            50 
ˆ
The force F  120 i  50 ˆ is at an angle   tan 1 
j                                   22 .6 from the line DC. Thus it does
 120 
not pass through O and intersects with side AD and diagonal BD of the square. Therefore:
(i)       It is not possible to keep the square in equilibrium by applying the third force at O.
(ii)      It is possible to keep the square in equilibrium by applying the third force at a point
on BD.
Equation of BD with O as origin is y  x

a 5   3a 
Equation of line along which the third force acts is y       x 
2 12   2 
3
Solving the two equations gives y  x        a
34
This gives the distance of point O = 0.125a
1 3 
And from B =     2    a  0.58 a
 2 34 
(iii)     It is clear that for equilibrium, the force can be applied only on AD and BC.

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 views: 12 posted: 7/13/2012 language: pages: 16