Calculate Load when the Effort is 25 kgf and mechanical advantage is 0.5.
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Explanation:
ANSWER
Given,
Effort = 25 kgf
Load = 75 kgf
As we know that,
M.A. = \dfrac{\text{Load}}{\text{Effort}} \\[0.5em]M.A.=
Effort
Load
Substituting the values in the formula we get,
M.A. = \dfrac{75}{25} \\[0.5em] M.A. = \dfrac{3}{1} \\[0.5em] \Rightarrow M.A. = 3 \\[0.5em]M.A.=
25
75
M.A.=
1
3
⇒M.A.=3
As we know,
V.R. = n = 3
\eta = \dfrac{M.A.}{V.R.} \\[0.5em] \eta = \dfrac{3}{3} \\[0.5em] \Rightarrow\eta = 1 \\[0.5em]η=
V.R.
M.A.
η=
3
3
⇒η=1
Therefore, efficiency = 100%
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