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Sunday, December 9, 2018

Chapter 5.15 - Solved examples on Force in Strings

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:
Free body diagram can be used to find tension in strings when objects are attached vertically
Fig.5.53
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 = FW=195ˆj150ˆj=45ˆj
• By Newton's second law, we have: Net force = ma
• So we have: 45ˆj=[(mA+mB+mC)×|a|]ˆj
|a|=4515=3ms2
• 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' = TRSWC
5. By Newton's second law, we have: Net force = ma
• So we have: TRSWC=(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
• 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' = TPQTRSWB
8. By Newton's second law, we have: Net force = ma
• So we have: TPQTRSWB=(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' = FTPQWA
By Newton's second law, we have: Net force = ma
• So we have: FTPQWA=(mA×|a|)ˆj
Substituting the values, we get: 195ˆj156ˆj30ˆj=(3×|a|)ˆj
|a|=3ms2
• 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
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=4ms2
• 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
• 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:
Fig.5.57
• 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' = TRSTPQ
8. By Newton's second law, we have: Net force = ma
• So we have: TRSTPQ=(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
• So net force acting on 'C' = FTRS
By Newton's second law, we have: Net force = ma
• So we have: FTRS=(mC×|a|)ˆj
Substituting the values, we get: 56ˆj48ˆj=(2×|a|)ˆj
|a|=4ms2
• 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
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:
F1+F2=0
    ♦ 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
• 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:
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|=503N

In the next section we will see strings through pulleys

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