The resultant of two forces P and Q (such that P > Q) acting along the same straight line, but in opposite direction, is given by
A. P + Q
B. P - Q
C. $$\frac{{\text{P}}}{{\text{Q}}}$$
D. $$\frac{{\text{Q}}}{{\text{P}}}$$
Answer: Option B
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
Join The Discussion
Comments ( 2 )
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
Taking summation of Fx= P-Q( p is positive coz mentioned in question), summation of Fy=0(no vertical forces). R= Fxsquare+ fy square under root. Here R=( (p-q)square - o square) under root. After cancellations we get R= P-Q.
Give reason about anwser