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.

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.

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)

Rahul kumar said:   1 decade ago
Rahul kumar.

F1+F2=20-------> (1).
14F1-6F2=20-----> (2).

Solving equation (1)& (2).

F1 = 5N.
F2 = 15N.

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.

M. Flowers said:   1 decade ago
I'm not understanding what method you applied to get these answers. What formula was used?

Nithyananda said:   10 years ago
They are given F1=24, but I don't know why the two person taken as F1=14?

Rajesh Kadam said:   5 years ago
F1 + F2 = 20
F1*24 = R * 18
SOLVING Eq we get;
F1 = 15.
F2 = 5.
(5)

Pandees waran said:   9 years ago
F1 + F2 = 20.

15 + 5 = 20 simple method.

Fathima said:   10 years ago
Can you give us a schematic sketch?

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


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