In the previous section, we saw center of gravity. In this section we will see some solved examples
Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support
Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
RA + RB - 6g - 4g = 0
⇒ RA + RB = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by RA about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by RB about A = RB × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5RB = 0
⇒ RB = 43.12 N
• Substituting this value of RB in (3), we get: RA = (98-43.12) = 54.88 N
Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:
Find the reaction forces on the wall and the floor. Take g = 9.8 ms-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as RA
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as RB
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C' • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
♦ The slipping is prevented entirely by the horizontal force F at A
♦ The vertical load 20 g is resisted entirely by the vertical force RA at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - RB = 0
(ii) In the y direction: RA - 20 g = 0
⇒ RA = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by RA about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by RB about A = RB × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise)
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ RB = 34.65 N
10. Substituting this value of RB in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• RA vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97o
Solved example 7.23
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:
The angles made by the strings with the vertical are 36.9o and 53.1o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T1 and T2 be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9o with the vertical wall
♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
♦ The angles are same because, they are alternate angles
• In fig.a, we see that, the right string makes an angle of 53.1o with the vertical wall
♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T1 sin θ1 + T2 sin θ2 = 0
⇒ T1 sin θ1 = T2 sin θ2.
⇒ T1 sin 36.9 = T2 sin 53.1 = T2 cos (90 - 36.9) = T2 cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T2 = 0.75 T1
(ii) In the y direction: T1 cos θ1 + T2 cos θ2 - W = 0
⇒ T1 cos 36.9 + T2 cos 53.1 = W
⇒0.8 T1 + 0.6 T2 = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T1 sin θ1 about G = zero
(ii) Torque created by T1 cos θ1 about G = T1 cos θ1 × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T2 cos θ2 about G = T2 cos θ2 × (2-d) (anti clockwise)
(i) Torque created by T2 sin θ2 about G = zero
4. Applying the condition, we get:
[T1 cos θ1 × d] - [T2 cos θ2 × (2-d)] = 0
⇒ [T1 × cos 36.9 × d] - [0.75 T1 × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T2 = 0.75 T1)
⇒ [0.8 T1 d] - [0.45 T1 (2-d)] = 0
⇒ 0.8 T1 d - 0.9 T1 + 0.45 T1 d = 0
⇒ 1.25 T1 d = 0.9 T1
⇒ 1.25 d = 0.9
⇒ d = 0.72 m
Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram
• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
RA + RB -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by RA about G = 1.05 RA (clockwise)
(ii) Torque created by 1800g about G = zero
(iii) Torque created by RB about G = 0.75 RB (anticlockwise)
4. Applying the condition, we get:
1.05 RA - 0.75 RB = 0
1.05 RA = 0.75 RB
RB = 1.4 RA
5. Substituting this in (2), we get:
RA + 1.4RA = 1800 × 9.8
⇒ RA = 7350 N
• So RB = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = 7350⁄2 = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = 10290⁄2 = 5145 N
Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms-2 [Hint: Consider the equilibrium of each side of the ladder separately]
Solution:
1. The detailed measurements are shown in fig.b
• Given:
♦ BD = 0.8 m
♦ DF = FA = 0.4 m
♦ DE = 0.5 m
• Let ∠ABC = θ
♦ Since DE is parallel to the ground, we get: ∠ADE = θ
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
♦ This is shown as red dashed line
♦ The foot of the perpendicular on BC is O
♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
♦ BD = 0.8 m
♦ DF = FA = 0.4 m
♦ DE = 0.5 m
♦ AG =0.76 m
♦ AO = 1.52 m
♦ BO = 0.5 m
♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:
• RB is the normal reaction at B
• RC is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• RB + RC - (40 × 9.8) = 0
⇒ RB + RC - 392 = 0
⇒ RB + RC = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by RB about O = RB × BO = RB × 0.5 = 0.5RB Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by RC about O = RC × CO = RC × 0.5 = 0.5RC Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5RB - 49 - 0.5RC = 0
⇒ 0.5RB - 0.5RC = 49
⇒ RB - RC = 98
11. Solving the equations in (8) and (10), we get:
RB = 245 N, RC = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: RB, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by RB about A = RB × BO = RB × 0.5 = 0.5RB Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5RB - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N
Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support
 |
| Fig.7.103 |
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
RA + RB - 6g - 4g = 0
⇒ RA + RB = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by RA about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by RB about A = RB × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5RB = 0
⇒ RB = 43.12 N
• Substituting this value of RB in (3), we get: RA = (98-43.12) = 54.88 N
Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:
 |
| Fig.7.104 |
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as RA
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as RB
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C' • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
♦ The slipping is prevented entirely by the horizontal force F at A
♦ The vertical load 20 g is resisted entirely by the vertical force RA at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - RB = 0
(ii) In the y direction: RA - 20 g = 0
⇒ RA = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by RA about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by RB about A = RB × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise)
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ RB = 34.65 N
10. Substituting this value of RB in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• RA vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97o
Solved example 7.23
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:
 |
| Fig.7.105 |
Solution:
1. The free body diagram is shown in fig.b
• Let T1 and T2 be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9o with the vertical wall
♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
♦ The angles are same because, they are alternate angles
• In fig.a, we see that, the right string makes an angle of 53.1o with the vertical wall
♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T1 sin θ1 + T2 sin θ2 = 0
⇒ T1 sin θ1 = T2 sin θ2.
⇒ T1 sin 36.9 = T2 sin 53.1 = T2 cos (90 - 36.9) = T2 cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T2 = 0.75 T1
(ii) In the y direction: T1 cos θ1 + T2 cos θ2 - W = 0
⇒ T1 cos 36.9 + T2 cos 53.1 = W
⇒0.8 T1 + 0.6 T2 = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T1 sin θ1 about G = zero
(ii) Torque created by T1 cos θ1 about G = T1 cos θ1 × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T2 cos θ2 about G = T2 cos θ2 × (2-d) (anti clockwise)
(i) Torque created by T2 sin θ2 about G = zero
4. Applying the condition, we get:
[T1 cos θ1 × d] - [T2 cos θ2 × (2-d)] = 0
⇒ [T1 × cos 36.9 × d] - [0.75 T1 × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T2 = 0.75 T1)
⇒ [0.8 T1 d] - [0.45 T1 (2-d)] = 0
⇒ 0.8 T1 d - 0.9 T1 + 0.45 T1 d = 0
⇒ 1.25 T1 d = 0.9 T1
⇒ 1.25 d = 0.9
⇒ d = 0.72 m
Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram
 |
| Fig.106 |
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
RA + RB -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by RA about G = 1.05 RA (clockwise)
(ii) Torque created by 1800g about G = zero
(iii) Torque created by RB about G = 0.75 RB (anticlockwise)
4. Applying the condition, we get:
1.05 RA - 0.75 RB = 0
1.05 RA = 0.75 RB
RB = 1.4 RA
5. Substituting this in (2), we get:
RA + 1.4RA = 1800 × 9.8
⇒ RA = 7350 N
• So RB = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = 7350⁄2 = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = 10290⁄2 = 5145 N
Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms-2 [Hint: Consider the equilibrium of each side of the ladder separately]
 |
| Fig.7.107 |
1. The detailed measurements are shown in fig.b
• Given:
♦ BD = 0.8 m
♦ DF = FA = 0.4 m
♦ DE = 0.5 m
• Let ∠ABC = θ
♦ Since DE is parallel to the ground, we get: ∠ADE = θ
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
♦ This is shown as red dashed line
♦ The foot of the perpendicular on BC is O
♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
♦ BD = 0.8 m
♦ DF = FA = 0.4 m
♦ DE = 0.5 m
♦ AG =0.76 m
♦ AO = 1.52 m
♦ BO = 0.5 m
♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:
 |
| Fig.7.107 (c) & (d) |
• RC is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• RB + RC - (40 × 9.8) = 0
⇒ RB + RC - 392 = 0
⇒ RB + RC = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by RB about O = RB × BO = RB × 0.5 = 0.5RB Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by RC about O = RC × CO = RC × 0.5 = 0.5RC Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5RB - 49 - 0.5RC = 0
⇒ 0.5RB - 0.5RC = 49
⇒ RB - RC = 98
11. Solving the equations in (8) and (10), we get:
RB = 245 N, RC = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: RB, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by RB about A = RB × BO = RB × 0.5 = 0.5RB Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5RB - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N
Two more solved examples can be seen here
In the next section, we will see moment of inertia
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