Question

What is the VR and Mechanical Advantage for the following pulley system (assuming that all cables are always in vertical alignment).

Answer 1

Hint:To solve this question, we need to understand the definitions of the mechanical advantage and the velocity ratio for a machine. The mechanical advantage is defined as the ratio of load to the effort. The velocity ratio is defined as the displacement of effort to the displacement of load. Here, we have given the simple pulley system for which we will determine both these parameters.

Formulas used:
$M.A. = \dfrac{L}{E}$,
where, $M.A.$ is the mechanical advantage, $L$ is the load and $E$ is the effort
$V.R. = \dfrac{{{d_E}}}{{{d_L}}}$,
where, $V.R.$ is the velocity ratio, ${d_E}$ is the displacement of effort and ${d_L}$ is the displacement of load
$\eta = \dfrac{{M.A.}}{{V.R.}}$,
where, $\eta $ is the efficiency of the machine, $M.A.$ is the mechanical advantage and $V.R.$ is the velocity ratio

Complete step by step answer:
Here, we are given the simple pulley system and we are also asked to assume that all the cables are always in vertical alignment.
In this case, if we consider the frictionless movement between the pulley and the cable and imagine the effort on the left hand rope pulled down a certain distance. This means that the same amount of length of rope must come over the pulley and so the load rises to the same distance.
We know that velocity ratio is given by
$V.R. = \dfrac{{{d_E}}}{{{d_L}}}$
Here, we have seen that ${d_E} = {d_L}$
Thus, the velocity ratio of the given system is 1.
Now for frictionless motion, the efficiency of a machine is $100\% $.
We know that
$\eta = \dfrac{{M.A.}}{{V.R.}}$
$ \Rightarrow M.A. = \eta \times V.R.$
Therefore, for the given case
$\therefore M.A. = 1 \times 1 = 1$
Therefore, for the given system, mechanical advantage is 1.

Thus, our final answer is: The value of both mechanical advantage and velocity ratio for the given system is 1.

Note:We have considered that the motion between the pulley and the cable is frictionless. This type of machine is called the ideal machine. For such ideal machines, work output is the same as the work input. This means that there are no energy losses and hence the efficiency will be $100\% $. However, in actual machine, there are always some energy losses due to motion of the moving parts which decreases its efficiency compared to that of the ideal machine,