In the previous section we saw the basic details about 'force in string'. We saw the possible 3 cases and saw a solved example related to case 3. In this section we will see some solved examples related to cases 1 and 2.
Solved example 5.24
Three blocks A, B and C are connected vertically as shown in fig.5.53(a) below:
Their masses are 3 kg, 5 kg and 7 kg respectively. If the system of blocks is raised upwards by a force of 195 N, what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.53(b) above
• In the FBD, there are 2 forces
• The weight W is the total weight of the 3 masses. So we can write:
• $\mathbf\small{|\vec{W}|=(m_A+m_B+m_C)g}$= 150 N
2. Net force acting on the system = $\mathbf\small{\vec{F}-\vec{W}=195\, \hat{j}-150\, \hat{j}=45\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{45}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{45}{15}=3\, \text{ms}^{-2}}$
• So the whole system moves upwards with an acceleration of 3 ms-2
3. Next we draw the FBD of 'C'. This is shown in fig.5.53(c) above
• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
4. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• So net force acting on 'C' = $\mathbf\small{\vec{T_{RS}}-\vec{W_C}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{RS}}-\vec{W_C}=(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=\vec{W_C}+(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=(m_C \times g)\hat{j}+(m_C \times |\vec{a}|})\hat{j}=[m_C \times (|\vec{a}|+g)]\hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{RS}|}=[m_C \times (|\vec{a}|+g)]}}$
• Substituting the values, we get:
$\mathbf\small{{|\vec{T_{RS}|}=[7 \times (3+10)]}=91\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.54(a) below:
• In the segment PQ, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
7. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• The tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = $\mathbf\small{\vec{T_{PQ}}-\vec{T_{RS}}-\vec{W_B}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{PQ}}-{\vec{T_{RS}}-\vec{W_B}=(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+\vec{W_B}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+(m_B \times g)\hat{j}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}=\vec{T_{RS}}+\left[m_B \times (|\vec{a}|+g) \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=|\vec{T_{RS}}|+\left[m_B \times (|\vec{a}|+g) \right ]}$
• Substituting the values, we get:
$\mathbf\small{|\vec{T_{PQ}}|=91+\left[5 \times (3+10) \right ]=156\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'A' is shown in fig.5.54(b). We have:
• So net force acting on 'A' = $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}=(m_A \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{195 \hat{j}-156 \hat{j}-30 \hat{j}=(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 3\, \text{ms}^{-2}}$
• We have already seen that each mass move upwards with an acceleration of 3 ms-2. So our calculations are correct
Solved example 5.25
Three blocks A, B and C connected horizontally rest on a friction less floor as shown in the fig.5.55(a) below:
Their masses are 7 kg, 5 kg and 2 kg respectively. If the system of blocks is moved towards the right by a force of 56 N applied on 'C', what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.55(b) above
• In the FBD, there is only 1 force
(The vertical forces will all cancel each other, and hence are not shown)
2. Net force acting on the system = $\mathbf\small{56\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{56}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{56}{14}=4\, \text{ms}^{-2}}$
• So the whole system moves towards the right with an acceleration of 4 ms-2
3. Next we draw the FBD of 'A'. This is shown in fig.5.56 below:
• In the segment PQ, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
4. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• So net force acting on 'A' = $\mathbf\small{\vec{T_{PQ}}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}=28\, \hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{PQ}}|=28}\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.57 below:
• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
7. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• So net force acting on 'B' = $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}=(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=\vec{T_{PQ}}+(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=28\, \hat{j}+20\, \hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{RS}}|=48\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'C' is shown in fig.5.58 below:
• So net force acting on 'C' = $\mathbf\small{\vec{F}-\vec{T_{RS}}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{RS}}=(m_C \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{56 \hat{j}-48 \hat{j}=(2 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 4\, \text{ms}^{-2}}$
• We have already seen that each mass move with an acceleration of 4 ms-2. So our calculations are correct
Now we will see a solved example based on case 3
Solved example 5.26
A block 'A' of mass 10 kg is suspended using 3 strings QP, QR and QS as shown in the fig.5.59(a) below. Find the tensions in the strings.
Solution:
1. First we draw the FBD of 'A'
• So 'A' is taken as the sub-system. A red rectangle is drawn around 'A' in fig.5.59(b)
• What ever is enclosed in the red rectangle, should be used in the FBD
• 'A' is in equilibrium under the action of two forces. So we apply the condition for equilibrium as follows:
• Direction of $\small{\vec{T_{QS}}}$ is obviously, 'vertical and upwards'
2. Now we draw the FBD of a sub-system around the point 'Q'
This is shown in fig.5.60 below:
• What ever is enclosed in the red rectangle, should be used in the FBD.
3. In fig.b, 'Q' is in equilibrium under the action of 3 forces. So we apply the condition for equilibrium as follows:
Solved example 5.24
Three blocks A, B and C are connected vertically as shown in fig.5.53(a) below:
 |
| Fig.5.53 |
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.53(b) above
• In the FBD, there are 2 forces
• The weight W is the total weight of the 3 masses. So we can write:
• $\mathbf\small{|\vec{W}|=(m_A+m_B+m_C)g}$= 150 N
2. Net force acting on the system = $\mathbf\small{\vec{F}-\vec{W}=195\, \hat{j}-150\, \hat{j}=45\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{45}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{45}{15}=3\, \text{ms}^{-2}}$
• So the whole system moves upwards with an acceleration of 3 ms-2
3. Next we draw the FBD of 'C'. This is shown in fig.5.53(c) above
• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
4. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• So net force acting on 'C' = $\mathbf\small{\vec{T_{RS}}-\vec{W_C}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{RS}}-\vec{W_C}=(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=\vec{W_C}+(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=(m_C \times g)\hat{j}+(m_C \times |\vec{a}|})\hat{j}=[m_C \times (|\vec{a}|+g)]\hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{RS}|}=[m_C \times (|\vec{a}|+g)]}}$
• Substituting the values, we get:
$\mathbf\small{{|\vec{T_{RS}|}=[7 \times (3+10)]}=91\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.54(a) below:
 |
| Fig.5.54 |
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
7. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• The tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = $\mathbf\small{\vec{T_{PQ}}-\vec{T_{RS}}-\vec{W_B}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{PQ}}-{\vec{T_{RS}}-\vec{W_B}=(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+\vec{W_B}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+(m_B \times g)\hat{j}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}=\vec{T_{RS}}+\left[m_B \times (|\vec{a}|+g) \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=|\vec{T_{RS}}|+\left[m_B \times (|\vec{a}|+g) \right ]}$
• Substituting the values, we get:
$\mathbf\small{|\vec{T_{PQ}}|=91+\left[5 \times (3+10) \right ]=156\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'A' is shown in fig.5.54(b). We have:
• So net force acting on 'A' = $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}=(m_A \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{195 \hat{j}-156 \hat{j}-30 \hat{j}=(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 3\, \text{ms}^{-2}}$
• We have already seen that each mass move upwards with an acceleration of 3 ms-2. So our calculations are correct
Solved example 5.25
Three blocks A, B and C connected horizontally rest on a friction less floor as shown in the fig.5.55(a) below:
 |
| Fig.5.55 |
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.55(b) above
• In the FBD, there is only 1 force
(The vertical forces will all cancel each other, and hence are not shown)
2. Net force acting on the system = $\mathbf\small{56\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{56}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{56}{14}=4\, \text{ms}^{-2}}$
• So the whole system moves towards the right with an acceleration of 4 ms-2
3. Next we draw the FBD of 'A'. This is shown in fig.5.56 below:
 |
| Fig.5.56 |
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
4. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• So net force acting on 'A' = $\mathbf\small{\vec{T_{PQ}}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}=28\, \hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{PQ}}|=28}\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.57 below:
 |
| Fig.5.57 |
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force'
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
7. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• So net force acting on 'B' = $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}=(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=\vec{T_{PQ}}+(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=28\, \hat{j}+20\, \hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{RS}}|=48\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'C' is shown in fig.5.58 below:
 |
| Fig.5.58 |
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{RS}}=(m_C \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{56 \hat{j}-48 \hat{j}=(2 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 4\, \text{ms}^{-2}}$
• We have already seen that each mass move with an acceleration of 4 ms-2. So our calculations are correct
Now we will see a solved example based on case 3
Solved example 5.26
A block 'A' of mass 10 kg is suspended using 3 strings QP, QR and QS as shown in the fig.5.59(a) below. Find the tensions in the strings.
 |
| Fig.5.59 |
1. First we draw the FBD of 'A'
• So 'A' is taken as the sub-system. A red rectangle is drawn around 'A' in fig.5.59(b)
• What ever is enclosed in the red rectangle, should be used in the FBD
• 'A' is in equilibrium under the action of two forces. So we apply the condition for equilibrium as follows:
$\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{QS}}}$
♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{W_A}}$
♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{QS}}}$
♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{W_A}}$
• Considering upward forces as positive and downward forces as negative, we get:
$\mathbf\small{(|\vec{T_{QS}}|)\hat{j}-(|\vec{W_A}|)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}-(m_A \times g)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}=(m_A \times g)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{QS}}|=m_A \times g= 10 \times 10 = 100\,N}$• Direction of $\small{\vec{T_{QS}}}$ is obviously, 'vertical and upwards'
2. Now we draw the FBD of a sub-system around the point 'Q'
This is shown in fig.5.60 below:
 |
| Fig.5.60 |
3. In fig.b, 'Q' is in equilibrium under the action of 3 forces. So we apply the condition for equilibrium as follows:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{PQ}}}$
♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{T_{QR}}}$
♦ Let $\small{\vec{F_3}}$ indicate $\small{\vec{T_{QS}}}$
■ Here two forces are inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
• Considering forces to the right as positive and forces to the left as negative, we get:
$\mathbf\small{-(|\vec{T_{PQ}}|\cos 60 )\hat{i}+(|\vec{T_{QR}}|\cos 30 )\hat{i}+\,0=0}$
(The last term is zero because, $\small{\vec{T_{QS}}}$ does not have a horizontal component)
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|\cos 60=|\vec{T_{QR}}|\cos 30}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
9. Considering the vertical components, we have:
$\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• Considering upward forces as positive and downward forces as negative, we get:
$\mathbf\small{(|\vec{T_{PQ}}|\sin 60 )\hat{j}+(|\vec{T_{QR}}|\sin 30 )\hat{j}-(|\vec{T_{QS}}|)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )\hat{j}=(100)\hat{j}}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )=100}$
$\mathbf\small{\Rightarrow \sqrt{3}\, |\vec{T_{PQ}}|+|\vec{T_{QR}}|=200}$
10. But from (8) we have: $\mathbf\small{|\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
• So the result in (9) becomes: $\mathbf\small{ \sqrt{3}\, (\sqrt{3}\,|\vec{T_{QR}}|)+|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow 4\,|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow |\vec{T_{QR}}|=50 \,\text{N}}$
11. So from (8) we get: $\mathbf\small{|\vec{T_{PQ}}|=50\sqrt{3}\,\text{N}}$
♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{PQ}}}$
♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{T_{QR}}}$
♦ Let $\small{\vec{F_3}}$ indicate $\small{\vec{T_{QS}}}$
■ Here two forces are inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
• Considering forces to the right as positive and forces to the left as negative, we get:
$\mathbf\small{-(|\vec{T_{PQ}}|\cos 60 )\hat{i}+(|\vec{T_{QR}}|\cos 30 )\hat{i}+\,0=0}$
(The last term is zero because, $\small{\vec{T_{QS}}}$ does not have a horizontal component)
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|\cos 60=|\vec{T_{QR}}|\cos 30}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
9. Considering the vertical components, we have:
$\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• Considering upward forces as positive and downward forces as negative, we get:
$\mathbf\small{(|\vec{T_{PQ}}|\sin 60 )\hat{j}+(|\vec{T_{QR}}|\sin 30 )\hat{j}-(|\vec{T_{QS}}|)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )\hat{j}=(100)\hat{j}}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )=100}$
$\mathbf\small{\Rightarrow \sqrt{3}\, |\vec{T_{PQ}}|+|\vec{T_{QR}}|=200}$
10. But from (8) we have: $\mathbf\small{|\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
• So the result in (9) becomes: $\mathbf\small{ \sqrt{3}\, (\sqrt{3}\,|\vec{T_{QR}}|)+|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow 4\,|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow |\vec{T_{QR}}|=50 \,\text{N}}$
11. So from (8) we get: $\mathbf\small{|\vec{T_{PQ}}|=50\sqrt{3}\,\text{N}}$
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