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 1 of 2.
Ankit Virus said:
1 year ago
R = Pcos45 + Psin45.
R = P(cos45 + sin45).
R = P(1/√2+1/√2) {from value cos45 and sin45=1/√2},
R = P(2/√2).
R = P√2.
Hence R = √2P.
R = P(cos45 + sin45).
R = P(1/√2+1/√2) {from value cos45 and sin45=1/√2},
R = P(2/√2).
R = P√2.
Hence R = √2P.
(4)
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)
Naveen said:
10 years ago
I did not understand this problem please help.
(2)
Philip said:
3 years ago
R*2 = P*2 + P*2 using Pythagoras theorem.
Let's make square root of both sides just to remove square from R.
R*((2)*(1/2)) = (P*2 + P*2)*(1/2).
R*(2/2) = (2P*2)*(1/2),
R = 2P*(2/2),
R = 2P.
Let's make square root of both sides just to remove square from R.
R*((2)*(1/2)) = (P*2 + P*2)*(1/2).
R*(2/2) = (2P*2)*(1/2),
R = 2P*(2/2),
R = 2P.
(2)
Kesav said:
4 years ago
Lamis theorem R/ sin 90 = P/sin 135.
i. e. R= √2 P.
i. e. R= √2 P.
(2)
Esakkiyappan said:
9 years ago
Law of a parallelogram.
We know that Resultant force R =Square of (P^2 + Q^2 +2PQcosangle value).
Here in the problem, The above figure shows the two equal forces at right angles acting at a point.
so angle value is 90 Degrees
R = Square of (P^2 + Q^2 + 2PQcos90), P = Q.
therefore R =Square of (P^2 + P^2 + 2P * P * 0).
R = Square of (2P^2 + 0).
R = Square of (2) * P.
Option C.
We know that Resultant force R =Square of (P^2 + Q^2 +2PQcosangle value).
Here in the problem, The above figure shows the two equal forces at right angles acting at a point.
so angle value is 90 Degrees
R = Square of (P^2 + Q^2 + 2PQcos90), P = Q.
therefore R =Square of (P^2 + P^2 + 2P * P * 0).
R = Square of (2P^2 + 0).
R = Square of (2) * P.
Option C.
(2)
Dileep valli kumar said:
9 years ago
R=√(p2+p2+ 2ppcos90).
Cos90 = 0.
ThenR = √[p2+p2].
R = √[2p2),
R = (√2)(p).
Cos90 = 0.
ThenR = √[p2+p2].
R = √[2p2),
R = (√2)(p).
(1)
Nayeem said:
7 years ago
This problem is about non coplanar forces. Why is everyone using parallelogram law of coplanar forces?
(1)
Deepak Kumar said:
7 years ago
Lami's theorem can be used only coplanar forces.
(1)
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.
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