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.
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)
Dr.. Ramesh B. R said:
1 decade ago
f1+f2 = 20 --->1.
Moment of the resultant about a point shall be equal to moment of the forces about the same point. Therefore take moment of the forces on a point abve resultant.
R*0 = F1*14-F2*6 ------>2.
Solving above two eqn.
F2 = 15N F1 = 5 N.
Moment of the resultant about a point shall be equal to moment of the forces about the same point. Therefore take moment of the forces on a point abve resultant.
R*0 = F1*14-F2*6 ------>2.
Solving above two eqn.
F2 = 15N F1 = 5 N.
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)
Pradeep said:
1 decade ago
Nothing the method is used here, It is simple, If a number of forces are applied on a body, then all the forces are replaced by a resultant force (algebraic sum of all forces). Here, there is no mean of distances given.
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.
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.
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)
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.
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)
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.
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.
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.
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