Question

The mechanical advantage of inclined plane of angle of inclination ${60^ \circ }$ is equal to:
(A) $\dfrac{2}{{\sqrt 3 }}$
(B) ${\text{cosec3}}{{\text{0}}^{\text{o}}}$
(C) Both (1) and (2)
(D) None of the above.

Answer 1

Hint: The mechanical advantage of the inclined plane is based on the principle of work which states that work output through the simple machine (the lever) is equal to the work input. Mechanical advantage of the inclined plane is the length of the slope divided by the height of the inclined plane.

Formula Used: The formulae used in the solution are given here.
Mechanical advantage of an inclined plane at an angle $\theta $ is, thus given by- $\dfrac{{Length}}{{Height}} = \dfrac{l}{h} = \dfrac{1}{{\sin \theta }}$

Complete step by step answer
The mechanical advantage of an inclined plane is the ratio of the load on the ramp over the effort required to pull upward the ramp. It depends on the inclination angle. The mechanical advantage of the inclined plane is based on the principle of work which states that work output through the simple machine (the lever)is equal to the work input. The work input is the load multiplied by the height while the work output through the machine is equal to the effort used multiplied by the length of the plane. This principle is true for any slope value.
Now, we know, work input = work output,
$ \Rightarrow $ load$ \times $ height = effort $ \times $ length
Thus, the ratio of load to effort is equal to the ratio of length to height of the inclined plane.
Thus, $\dfrac{{Load}}{{Effort}} = \dfrac{{Length}}{{Height}}$.
In an inclined plane, the formula for mechanical advantage is the length of the slope divided by the height of the inclined plane.
Mechanical advantage of an inclined plane at an angle $\theta $ is, thus given by- $\dfrac{{Length}}{{Height}} = \dfrac{l}{h} = \dfrac{1}{{\sin \theta }}$
If the inclination angle is zero degrees with respect to the horizontal then it is not an inclined plane any longer.
If the angle of inclination is ${60^ \circ }$ with respect to the horizontal, then the height is equal to $0.87$ times the length. The mechanical advantage at this angle is $1/0.87$ or about only $1.15$. This means that the inclined plane had lessened your force by about $87$ percent of the weight of the object. The inclined plane reduces the effort by $0.87$ and lengthens the distance, the object moves by $1/0.87$ or $1.15$.
Given that, the angle of inclination is ${60^ \circ }$. From the equation given above,
Mechanical advantage of an inclined plane at an angle $\theta $ is, thus given by- $\dfrac{{Length}}{{Height}} = \dfrac{l}{h} = \dfrac{1}{{\sin \theta }}$
where $\theta = {60^ \circ }$.
Thus the mechanical advantage is given by, $\dfrac{1}{{\sin {{60}^ \circ }}} = {\text{cosec 6}}{{\text{0}}^{\text{o}}} = \dfrac{2}{{\sqrt 3 }}.$.
The mechanical advantage of an inclined plane of angle of inclination ${60^ \circ }$ is equal to $\dfrac{2}{{\sqrt 3 }}$.

The correct answer is Option A.

Note
If the inclination angle is ${30^ \circ }$ with respect to the horizontal then the height is equal to one-half of its length. The mechanical advantage at this angle is $2$. This means that the force needed to raise the object from the bottom of the plane is one-half of the object’s weight. The inclined plane reduces the effort by two and lengthens the distance the object moves by $2$.
At an inclination angle of ${90^ \circ }$ the height is equal to the length and the MA is equal to $1$. This means that it is useless to use the inclined plane because the force needed to raise the object is equal to the weight of the object. It is better to lift the object directly without the inclined plane.