Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 49)
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.

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.

Lova said:   9 years ago
Thank you very much friends.

Rahul s said:   8 years ago
Thank you very much @Pandees Waran.

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.

Prashant Ranjan said:   8 years ago
F1 !---6mm--*-----18mm--------! F2.

Anti-clockwise moment = Clockwise moment
F1*6 = F2*18
F1 = 3F2--------------(1)
F1+F2 = 20 ( given)
F1 = 20-F2------------(2)
From (1) and (2), we get
3F2 = 20-F2.
F2 = 5 kn.
F1 = 3F2 =3*5=15kn.
(2)

Nikss said:   8 years ago
@ALL.

Two forces acting parallel in the same direction. So the resultant will be a single force in the same direction. So from where this clockwise and anticlockwise concept came?

All forces will be in same direction.

Faisal said:   8 years ago
Thanks @Roj.
(1)

Kaku said:   7 years ago
F1 + F2 = 20N.

Taking moment about B.
R*0= -f1(6)+ f2(24-6),
0 = -(20-f2)(6) + 18f2,
F2 =5.
After putting we will get the answer.
(3)

Sanjeev said:   7 years ago
Thanks @Roj.

Praveen said:   6 years ago
This is to be solved by Varignon's Theorem.

Consider a point A.

Moment of F1 about A + Moment of F2 about A = Moment of R about A.


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