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The resultant of two equal forces P making an angle $$\theta ,$$ is given by
Answer & Solution
Correct Answer:
Option
B
To find the resultant of two equal forces P making an angle θ, we use the formula for the resultant of two forces:
Resultant $$R$$ = $$2{\text{P}}\cos \frac{\theta }{2}$$
Here’s the derivation:
- The two forces P are equal in magnitude and form an angle θ with each other.
- The magnitude of the resultant force is given by $$2{\text{P}}\cos \frac{\theta }{2}$$ where $$ \frac{\theta }{2}$$ is the angle between one of the forces and the resultant.
The other options involve sine, tangent, or cotangent functions, which do not correctly represent the magnitude of the resultant in this context.
Resultant $$R$$ = $$2{\text{P}}\cos \frac{\theta }{2}$$
Here’s the derivation:
- The two forces P are equal in magnitude and form an angle θ with each other.
- The magnitude of the resultant force is given by $$2{\text{P}}\cos \frac{\theta }{2}$$ where $$ \frac{\theta }{2}$$ is the angle between one of the forces and the resultant.
The other options involve sine, tangent, or cotangent functions, which do not correctly represent the magnitude of the resultant in this context.
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Loginaccording to the parallelogram law , (which is made for a two equal concurrent forces). there is a formula for a find out the value of a resultant (R). and according to that formula . theta = 2pcos(theta/2)
F1 = F2 = P
Resultant Force = R
R = √[ (F1)^2 + (F2)^2 + 2*F1*F2*cosx ]
R = √[ P^2 + P^2 + 2*P*P*cosx ]
R = √[ 2P^2 + 2P^2*cosx ]
R = √[ 2P^2 ( 1 + cosx ) ]
R = √[ 2P^2 *( 2 {cos[x/2]}^2 ) ]
R = √[ 4P^2 * {cos[x/2]}^2 ]
R = 2*P*cos[x/2]
R^2=P^+Q^2+2PQcos(theta)
R^2=2P^2[1+cos(theta)]
R^2=2P^2[2cos^2(theta/2)]
R^2=4P^2cos^2(theta/2)
R^2=2P cosθ/2