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
P/2
2P
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.
Sajal said:
1 decade ago
R = root of (p2+p2+2p2 cos 90).
= root of (2p2).
= root 2*p.
= root of (2p2).
= 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.
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.
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