JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 2/ Question on Equilibrium of forces solved by resolution of forces into rectangular components

1. Physics Helpline
L K Satapathy Laws of Motion 2
Equilibrium of a System of Forces

2. Physics Helpline
L K Satapathy Laws of Motion 2
Question : A block of mass 10 kg is
suspended from a horizontal bar
using two strings OA and OB of
lengths 9m and 12m respectively. The
ends A and B of the strings are fixed
at two points on the bar 15m apart as
shown in the figure. The tension in
the string OA is [ Take g = 10 m/s ]
(a) 100 N (b) 80 N (c) 60 N (d) 40 N
10 kg
15 m
12 m9 m
1T 2T
A B
O
2

3. Physics Helpline
L K Satapathy Laws of Motion 2
Answer :
1 2
9
sin cos 0.6
15
OA
AB
    
2 1
12
sin cos 0.8
15
OB
AB
    
Components of T1 :
Given : OA = 9m OB = 12m AB = 15m
2 2 2
90o
AB OA OB AOB    
In  AOB :
1 1 1 1( ) sin ( ) cosi T along OX ii T along OD 
Components of T2 : 2 2 2 2( ) sin ( ) cosiii T along OX iv T along OD 
1 2 1 2, 90o
AOD BOD         
A
1T
2T
B
O
XX 
W
D
1
1 2
2
2 2sinT 1 1sinT 
1 1cosT 
2 2cosT 

4. Physics Helpline
L K Satapathy Laws of Motion 2
1 1 2 2sin sinT T 
1 1 2 2cos cosT T W  
1 2 1 2
4
0.6 0.8 . . . (1)
3
T T T T     
1 20.8 0.6 100T T Mg     
1 24 3 500 . . . (2)T T  
2 2 2
4 25
(1) (2) 4 3 500 500
3 3
and T T T     
2 1 2
4
60 80
3
T N T T N    
Horizontal equilibrium 
Vertical equilibrium 
2 2sinT 1 1sinT 
W
1 1cosT 
2 2cosT 
Correct option = (b)

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L K Satapathy
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