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.
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.
Arshad said:
8 years ago
R^2 = P^2 + P^2 + 2P^2cos90 =>
R^2 = 2P^2 + 0 =>
R = √2P.
R^2 = 2P^2 + 0 =>
R = √2P.
(4)
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.
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.
Kesav said:
4 years ago
Lamis theorem R/ sin 90 = P/sin 135.
i. e. R= √2 P.
i. e. R= √2 P.
(2)
Deepak Kumar said:
7 years ago
Lami's theorem can be used only coplanar forces.
(1)
Naveen said:
10 years ago
I did not understand this problem please help.
(2)
Mrityunjay said:
1 decade ago
R = root(p^2+p^2).
= root(2p^2).
= root(2p).
= root(2p^2).
= root(2p).
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