The velocity ratio in case of an inclined plane inclined at angle 'θ' to the horizontal and weight being pulled up the inclined plane by vertical effort is
A. sinθ
B. cosθ
C. tanθ
D. cosecθ
Answer: Option A
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
Join The Discussion
Comments ( 4 )
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
Velocity Ratio = Effort application distance/Load raised distance
For this problem effort is applied vertically i.e. effort application distance is the perpendicular length of the triangle whose hypotenuse is the length of the inclined plane (making theta angle with horizontal). Load is raised through that length of the hypotenuse. Thus,
Perpendicular distance / hypotenuse will give sin(theta).
use the resolution of forces method.
an inclined line can be resolved into 2 components.
If weight is resolved w sin θ acts parallel to plane
Then
P= w sinθ
MA = W/P = 1/ sin θ = Cosecθ
So answer is d
Explain