Question

In a lifting machine and effort of 40 n raised a load of 1 kn .If efficiency of machine is 0.5 what is velocity ratio

Answer 1

Explanation:

Concept:

Mechanical Advantage: In a simple machine when the effort(P) balances a load (W), the ration of the load to the effort is called MA.

\(M.A = \frac{{Load}}{{Effort}} = \frac{W}{P}\)

Velocity Ratio: It is the ratio between the distance moved by the effort to the distance moved by the load.

\(V.R = \frac{{distance\;moved\;by\;the\;effort\;\left( {{d_P}} \right)}}{{distance\;moved\;by\;the\;load\;\left( {{d_w}} \right)}}\)

Efficiency: It is the ratio of output to input. In a simple mechanism, it is also defined as the ratio of mechanical advantage to the velocity ratio.

\(\eta = \frac{{Output}}{{Input}} = \frac{{M.A}}{{V.R}}\)

Calculation:

Given: Load, W = 1 kN = 1000 N, Effort, P = 40 N, η = 0.5

V.R. = ?

\(M.A = \frac{W}{P}= \frac{1000}{40}=25\)

\(\eta = \frac{{M.A}}{{V.R}}\Rightarrow 0.5 = \frac{{25}}{{V.R}} \Rightarrow V.R = 50\)

Answer 2

Answer:

The velocity ratio of the lifting machine is 50.

Explanation:

Mechanical Advantage:

In a simple machine when the effort(P) balances a load (W), the ratio of the load to the effort exists named MA.

$\(M.A = \frac{{Load}}{{Effort}} = \frac{W}{P}\)$

Velocity Ratio:

It exists the ratio between the distance moved by the effort to the distance moved by the load.

$\(V.R = \frac{{distance\;moved\;by\;the\;effort\;\left( {{d_P}} \right)}}{{distance\;moved\;by\;the\;load\;\left( {{d_w}} \right)}}\)$

Efficiency:

It exists the ratio of output to input. In a simple mechanism, it exists also described as the ratio of mechanical advantage to the velocity ratio.

Given:

Load, W = 1 kN = 1000 N Effort,

P = 40 n,

η = 0.5

velocity ratio =?

$\(M.A = \frac{W}{P}= \frac{1000}{40}=25\)$

$\(\eta = \frac{{M.A}}{{V.R}}$

$\Rightarrow 0.5 = \frac{{25}}{{V.R}}$

$\Rightarrow V.R = 50\)$

velocity ratio = 50.

Therefore, the velocity ratio of the lifting machine is 50.

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