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:
• |→W|=(mA+mB+mC)g= 150 N
2. Net force acting on the system = →F−→W=195ˆj−150ˆj=45ˆj
• By Newton's second law, we have: Net force = ma
• So we have: 45ˆj=[(mA+mB+mC)×|→a|]ˆj
⇒|→a|=4515=3ms−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 →TRS
• So net force acting on 'C' = →TRS−→WC
5. By Newton's second law, we have: Net force = ma
• So we have: →TRS−→WC=(mC×|→a|)ˆj
⇒→TRS=→WC+(mC×|→a|)ˆj
⇒→TRS=(mC×g)ˆj+(mC×|→a|)ˆj=[mC×(|→a|+g)]ˆj
⇒|→TRS|=[mC×(|→a|+g)]
• Substituting the values, we get:
|→TRS|=[7×(3+10)]=91N
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 →TPQ
• The tension in segment RS is denoted as →TRS
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = →TPQ−→TRS−→WB
8. By Newton's second law, we have: Net force = ma
• So we have: →TPQ−→TRS−→WB=(mB×|→a|)ˆj
⇒→TPQ=→TRS+→WB+(mB×|→a|)ˆj
⇒→TPQ=→TRS+(mB×g)ˆj+(mB×|→a|)ˆj
⇒→TPQ=→TRS+[mB×(|→a|+g)]ˆj
⇒|→TPQ|=|→TRS|+[mB×(|→a|+g)]
• Substituting the values, we get:
|→TPQ|=91+[5×(3+10)]=156N
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' = →F−→TPQ−→WA
By Newton's second law, we have: Net force = ma
• So we have: →F−→TPQ−→WA=(mA×|→a|)ˆj
Substituting the values, we get: 195ˆj−156ˆj−30ˆj=(3×|→a|)ˆj
⇒|→a|=3ms−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 = 56ˆj
• By Newton's second law, we have: Net force = ma
• So we have: 56ˆj=[(mA+mB+mC)×|→a|]ˆj
⇒|→a|=5614=4ms−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 →TPQ
• So net force acting on 'A' = →TPQ
5. By Newton's second law, we have: Net force = ma
• So we have: →TPQ=(mA×|→a|)ˆj
⇒→TPQ=(mA×|→a|)ˆj=28ˆj
⇒|→TPQ|=28N
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 →TRS
• So net force acting on 'B' = →TRS−→TPQ
8. By Newton's second law, we have: Net force = ma
• So we have: →TRS−→TPQ=(mB×|→a|)ˆj
⇒→TRS=→TPQ+(mB×|→a|)ˆj
⇒→TRS=28ˆj+20ˆj
⇒|→TRS|=48N
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' = →F−→TRS
By Newton's second law, we have: Net force = ma
• So we have: →F−→TRS=(mC×|→a|)ˆj
Substituting the values, we get: 56ˆj−48ˆj=(2×|→a|)ˆj
⇒|→a|=4ms−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 →TQS 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:
• |→W|=(mA+mB+mC)g= 150 N
2. Net force acting on the system = →F−→W=195ˆj−150ˆj=45ˆj
• By Newton's second law, we have: Net force = ma
• So we have: 45ˆj=[(mA+mB+mC)×|→a|]ˆj
⇒|→a|=4515=3ms−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 →TRS
• So net force acting on 'C' = →TRS−→WC
5. By Newton's second law, we have: Net force = ma
• So we have: →TRS−→WC=(mC×|→a|)ˆj
⇒→TRS=→WC+(mC×|→a|)ˆj
⇒→TRS=(mC×g)ˆj+(mC×|→a|)ˆj=[mC×(|→a|+g)]ˆj
⇒|→TRS|=[mC×(|→a|+g)]
• Substituting the values, we get:
|→TRS|=[7×(3+10)]=91N
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 →TPQ
• The tension in segment RS is denoted as →TRS
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = →TPQ−→TRS−→WB
8. By Newton's second law, we have: Net force = ma
• So we have: →TPQ−→TRS−→WB=(mB×|→a|)ˆj
⇒→TPQ=→TRS+→WB+(mB×|→a|)ˆj
⇒→TPQ=→TRS+(mB×g)ˆj+(mB×|→a|)ˆj
⇒→TPQ=→TRS+[mB×(|→a|+g)]ˆj
⇒|→TPQ|=|→TRS|+[mB×(|→a|+g)]
• Substituting the values, we get:
|→TPQ|=91+[5×(3+10)]=156N
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' = →F−→TPQ−→WA
By Newton's second law, we have: Net force = ma
• So we have: →F−→TPQ−→WA=(mA×|→a|)ˆj
Substituting the values, we get: 195ˆj−156ˆj−30ˆj=(3×|→a|)ˆj
⇒|→a|=3ms−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 = 56ˆj
• By Newton's second law, we have: Net force = ma
• So we have: 56ˆj=[(mA+mB+mC)×|→a|]ˆj
⇒|→a|=5614=4ms−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 →TPQ
• So net force acting on 'A' = →TPQ
5. By Newton's second law, we have: Net force = ma
• So we have: →TPQ=(mA×|→a|)ˆj
⇒→TPQ=(mA×|→a|)ˆj=28ˆj
⇒|→TPQ|=28N
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 →TRS
• So net force acting on 'B' = →TRS−→TPQ
8. By Newton's second law, we have: Net force = ma
• So we have: →TRS−→TPQ=(mB×|→a|)ˆj
⇒→TRS=→TPQ+(mB×|→a|)ˆj
⇒→TRS=28ˆj+20ˆj
⇒|→TRS|=48N
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: →F−→TRS=(mC×|→a|)ˆj
Substituting the values, we get: 56ˆj−48ˆj=(2×|→a|)ˆj
⇒|→a|=4ms−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:
→F1+→F2=0
♦ Let →F1 indicate →TQS
♦ Let →F2 indicate →WA
♦ Let →F1 indicate →TQS
♦ Let →F2 indicate →WA
• Considering upward forces as positive and downward forces as negative, we get:
(|→TQS|)ˆj−(|→WA|)ˆj=0
⇒(|→TQS|)ˆj−(mA×g)ˆj=0
⇒(|→TQS|)ˆj=(mA×g)ˆj
⇒|→TQS|=mA×g=10×10=100N• Direction of →TQS 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:
→F1+→F2+→F3=0
♦ Let →F1 indicate →TPQ
♦ Let →F2 indicate →TQR
♦ Let →F3 indicate →TQS
■ Here two forces are inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
→F1x+→F2x+→F3x=0
• Considering forces to the right as positive and forces to the left as negative, we get:
−(|→TPQ|cos60)ˆi+(|→TQR|cos30)ˆi+0=0
(The last term is zero because, →TQS does not have a horizontal component)
⇒|→TPQ|cos60=|→TQR|cos30
⇒|→TPQ|=√3|→TQR|
9. Considering the vertical components, we have:
→F1y+→F2y+→F3y=0
• Considering upward forces as positive and downward forces as negative, we get:
(|→TPQ|sin60)ˆj+(|→TQR|sin30)ˆj−(|→TQS|)ˆj=0
⇒(|→TPQ|sin60+|→TQR|sin30)ˆj=(100)ˆj
⇒(|→TPQ|sin60+|→TQR|sin30)=100
⇒√3|→TPQ|+|→TQR|=200
10. But from (8) we have: |→TPQ|=√3|→TQR|
• So the result in (9) becomes: √3(√3|→TQR|)+|→TQR|=200
⇒4|→TQR|=200
⇒|→TQR|=50N
11. So from (8) we get: |→TPQ|=50√3N
♦ Let →F1 indicate →TPQ
♦ Let →F2 indicate →TQR
♦ Let →F3 indicate →TQS
■ Here two forces are inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
→F1x+→F2x+→F3x=0
• Considering forces to the right as positive and forces to the left as negative, we get:
−(|→TPQ|cos60)ˆi+(|→TQR|cos30)ˆi+0=0
(The last term is zero because, →TQS does not have a horizontal component)
⇒|→TPQ|cos60=|→TQR|cos30
⇒|→TPQ|=√3|→TQR|
9. Considering the vertical components, we have:
→F1y+→F2y+→F3y=0
• Considering upward forces as positive and downward forces as negative, we get:
(|→TPQ|sin60)ˆj+(|→TQR|sin30)ˆj−(|→TQS|)ˆj=0
⇒(|→TPQ|sin60+|→TQR|sin30)ˆj=(100)ˆj
⇒(|→TPQ|sin60+|→TQR|sin30)=100
⇒√3|→TPQ|+|→TQR|=200
10. But from (8) we have: |→TPQ|=√3|→TQR|
• So the result in (9) becomes: √3(√3|→TQR|)+|→TQR|=200
⇒4|→TQR|=200
⇒|→TQR|=50N
11. So from (8) we get: |→TPQ|=50√3N
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