Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 20)
20.

The above figure shows the two equal forces at right angles acting at a point. The value of force R acting along their bisector and in opposite direction is
Discussion:
18 comments Page 2 of 2.
K.Tamil selvan said:
10 years ago
Resolving the forces:
ΓH=p.
ΓV=p .
Resultant force:
R = sq.root of ((ΓH^2)+(ΓV^2)).
R = sq.root ((p^2)+(p^2)).
R = sq.root (2p^2).
Ans: R = sq.root(2)*p.
ΓH=p.
ΓV=p .
Resultant force:
R = sq.root of ((ΓH^2)+(ΓV^2)).
R = sq.root ((p^2)+(p^2)).
R = sq.root (2p^2).
Ans: R = sq.root(2)*p.
BINIT RANJAN said:
10 years ago
Use Lami's theorem to get the answer in easy way, angle between two equal forces is 90 and that of bisector and any of two force is 135.
Happyhari25 said:
1 decade ago
R = ROOT(P^2+P^2) = ROOT(2P^2) = ROOT(2)P.
Therefore (C). ROOT(2)P.
Therefore (C). ROOT(2)P.
SHRIPAL said:
1 decade ago
R = Psin45+Pcos45.
R = P(sin45+cos45).
R = P*(2*root of 2/2).
R = P*root of 2.
R = P(sin45+cos45).
R = P*(2*root of 2/2).
R = P*root of 2.
Mrityunjay said:
1 decade ago
R = root(p^2+p^2).
= root(2p^2).
= root(2p).
= root(2p^2).
= root(2p).
Sakthi said:
1 decade ago
How can we conclude that "R"is the resultant force? If we use Law of sines, I get a different answer.
A.sindhuja said:
1 decade ago
Generally R^2=A^2+B^2+2ABcosX if A, B are the forces, R is the resultant and X is the angle between them.
Here P, P are the forces and they are perpendicular to each other then X=90 hence R^2=P^2+P^2+2*P*P*cos90 we know that cos90=0.
R^2 = P^2+P^2+0.
R^2 = 2P^2.
R = root of 2*p.
Here P, P are the forces and they are perpendicular to each other then X=90 hence R^2=P^2+P^2+2*P*P*cos90 we know that cos90=0.
R^2 = P^2+P^2+0.
R^2 = 2P^2.
R = root of 2*p.
1718 said:
1 decade ago
Resultant R = square of p+square of p.
Square of R = 2*square of p.
R = ROOT OF 2 *P.
Square of R = 2*square of p.
R = ROOT OF 2 *P.
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