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.
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)
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.
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)
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.
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)
Nayeem said:
7 years ago
This problem is about non coplanar forces. Why is everyone using parallelogram law of coplanar forces?
(1)
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.
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)
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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