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Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 49)
Engineering Mechanics
49.
two like parallel forces are acting at a distance of 24 mm apart and their resultant is 20 N. It the line of action of the resultant is 6 mm from any given force, the two forces are
15 N and 5 N
20 N and 5 N
15 N and 15 N
none of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
24 comments Page 2 of 3.
KRISH REDDY said:   8 years ago
Simple method friends.

All forces equal to the resultant is 20.

So see the option s where the 2 forces are 20 so that will be the answer. 5+15=20.
Rahul s said:   8 years ago
Thank you very much @Pandees Waran.
Lova said:   9 years ago
Thank you very much friends.
POPPY said:   9 years ago
Resultant R = F1 + F2 = 20N (i) F1----6mm-----R------18mm------------F2 equating clockwise moment to anticlockwise moment then, 18F2 = 6F1 THEN, F1 = 3F2, From equating in (i) we get F1=15 &F2 = 5.
Pandees waran said:   9 years ago
F1 + F2 = 20.

15 + 5 = 20 simple method.
ROJ said:   10 years ago
F1+F2 = 20.......1

Here 24 mm means total distance.

Take moment about resultant.

Note that moment = force*moment arm.

R*0 = F1*(24-6)-F2(6).
0 = 18F1-6F2.

18F1 = 6F2.
F1/F2 = 6/18 = 1:3.

So F1 = 5 & F2 = 15.
(1)
Nithyananda said:   10 years ago
They are given F1=24, but I don't know why the two person taken as F1=14?
Fathima said:   10 years ago
Can you give us a schematic sketch?
Rishi said:   1 decade ago
Where did 18f1 and 6f2 came from? I think you can simply add the forces as they are like and parallel. F1+F2=20 and option 1 satisfies it.
DARSHAN said:   1 decade ago
F1+F2=20......(1).

NOW,
MOMENT about Resultant
18F1-6F2=0.....(2).

By solving above eq.
We get F1=5 & F2=15.
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