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 3 of 3.
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.
NOW,
MOMENT about Resultant
18F1-6F2=0.....(2).
By solving above eq.
We get F1=5 & F2=15.
Rahul kumar said:
1 decade ago
Rahul kumar.
F1+F2=20-------> (1).
14F1-6F2=20-----> (2).
Solving equation (1)& (2).
F1 = 5N.
F2 = 15N.
F1+F2=20-------> (1).
14F1-6F2=20-----> (2).
Solving equation (1)& (2).
F1 = 5N.
F2 = 15N.
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.
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.
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