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
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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