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
Discussion:
24 comments Page 1 of 3.
Rajesh Kadam said:
5 years ago
F1 + F2 = 20
F1*24 = R * 18
SOLVING Eq we get;
F1 = 15.
F2 = 5.
F1*24 = R * 18
SOLVING Eq we get;
F1 = 15.
F2 = 5.
(5)
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.
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)
Suman said:
4 years ago
Here, the Resultant is 20N, So a summation of F1 & F2 will be also 20N.
Hence, Only option (a) is suitable as its sum of two numbers (15+5) is equal to 20N.
Hence, Only option (a) is suitable as its sum of two numbers (15+5) is equal to 20N.
(2)
Sujit said:
5 years ago
Let they are P &Q apart 24 mm & Assume Resultant acts at 6mm from left force means 18mm from right side (18+6=24mm).
Now we get moment about Resultant i.e. 6mm from left &18mm from right we get eq.
6P-18Q=0-> eq.1; i.e(submation of moment=0).
P+ Q = 20-> eq.2; i.e ( Resultant is 20).
After solving the above eq. We get value 15&5N.
Now we get moment about Resultant i.e. 6mm from left &18mm from right we get eq.
6P-18Q=0-> eq.1; i.e(submation of moment=0).
P+ Q = 20-> eq.2; i.e ( Resultant is 20).
After solving the above eq. We get value 15&5N.
(2)
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.
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)
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.
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)
Faisal said:
8 years ago
Thanks @Roj.
(1)
Rahul s said:
8 years ago
Thank you very much @Pandees Waran.
Arpan Acharya said:
5 years ago
Thank you all for explaining.
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.
Consider a point A.
Moment of F1 about A + Moment of F2 about A = Moment of R about A.
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