Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 49)
49.
If two equal forces of magnitude P act at an angle θ, their resultant, will be
Discussion:
14 comments Page 1 of 2.
Sofeeya Mulla said:
4 years ago
Thanks @Ankita.
Engineers study said:
5 years ago
We know,
Cosx = cos^2(x/2) - sin^2(x/2).
Adding 1 both side,
1+ cosx = cos^2(x/2) - sin^2(x/2)+1.
We also know,
sin^2(x/2)+cos^2(x/2) = 1.
So,
1+ cosx = cos^2(x/2) - sin^2(x/2)+sin^2(x/2)+cos^2(x/2).
1+ cosx = 2cos^2(x/2).
Apply the value of 1+ cosx and proceed. Ankitsinh Rajput thanks, dude.
Cosx = cos^2(x/2) - sin^2(x/2).
Adding 1 both side,
1+ cosx = cos^2(x/2) - sin^2(x/2)+1.
We also know,
sin^2(x/2)+cos^2(x/2) = 1.
So,
1+ cosx = cos^2(x/2) - sin^2(x/2)+sin^2(x/2)+cos^2(x/2).
1+ cosx = 2cos^2(x/2).
Apply the value of 1+ cosx and proceed. Ankitsinh Rajput thanks, dude.
(1)
Mukesh joshi said:
6 years ago
Thanks @Ankitsinh.
Ayush said:
6 years ago
Thanks all for explaining it.
Bitam said:
6 years ago
Thanks for explaining @Ankitsin.
Lakhan Bhavnani said:
6 years ago
Thanks for the explanation @Ankitsinh.
Abdul said:
7 years ago
@Ankitsinh.
Thanks for the explanation.
Thanks for the explanation.
Prathibha said:
7 years ago
Thanks for the explanation @Ankitsinh.
K.srinu said:
7 years ago
Thanks @Ankitsinh.
Ankitsinh Rajput said:
7 years ago
Here is the answer.
If two forces P and Q acting at angle θ then resultant R=(P^2+Q^2+2P*Q*cosθ)^1/2.
now forces are equal i.e P=Q.
R=(2P^2+2P^2cosθ)^1/2
R=P{2(1+cosθ)}^1/2
now , 1+cosθ=2cos^2(θ/2).
R=P{4cos^2(θθ/2)}^1/2.
R=2Pcos(θ/2) i.e ANS.
If two forces P and Q acting at angle θ then resultant R=(P^2+Q^2+2P*Q*cosθ)^1/2.
now forces are equal i.e P=Q.
R=(2P^2+2P^2cosθ)^1/2
R=P{2(1+cosθ)}^1/2
now , 1+cosθ=2cos^2(θ/2).
R=P{4cos^2(θθ/2)}^1/2.
R=2Pcos(θ/2) i.e ANS.
(2)
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