The resultant of two equal forces P making an angle $$\theta ,$$ is given by
A. $$2{\text{P}}\sin \frac{\theta }{2}$$
B. $$2{\text{P}}\cos \frac{\theta }{2}$$
C. $$2{\text{P}}\tan \frac{\theta }{2}$$
D. $$2{\text{P}}\cot \frac{\theta }{2}$$
Answer: Option B
Solution (By Examveda Team)
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.
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
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Comments (9)
The resultant of two equal forces P making an angle $$\theta ,$$ is given by
A. $$2{\text{P}}\sin \frac{\theta }{2}$$
B. $$2{\text{P}}\cos \frac{\theta }{2}$$
C. $$2{\text{P}}\tan \frac{\theta }{2}$$
D. $$2{\text{P}}\cot \frac{\theta }{2}$$
A. Equal to
B. Less than
C. Greater than
D. None of these
If a number of forces are acting at a point, their resultant is given by
A. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2}$$
B. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2}} $$
C. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)$$
D. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)} $$
A. $${\text{a}} = \frac{\alpha }{{\text{r}}}$$
B. $${\text{a}} = \alpha {\text{r}}$$
C. $${\text{a}} = \frac{{\text{r}}}{\alpha }$$
D. None of these
THE RESUILT OF TWO EQUAL FORCE "P" MAKING AN ANGLE
Work done per cycle is calculated as-
here is the ans,
according 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]
https://youtu.be/6dCrE6XxGzw
why the resultant of two equal forces 'P' making an angle theta is given by (2P COS theta/2) why not 2P sin theta/2
law of parallelogram is used that is
R^2=P^+Q^2+2PQcos(theta)
R^2=P^2+P^2+2P^2cos(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
(1 + cost) = 2 cos square t by 2
dgh