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A pulley system with V.R. = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction. (g = 10 N kg-1)

Calculate:

(i) Distance moved by the effort.

(ii) Work done by the effort.

(iii) M.A. of the pulley system.

(iv) Efficiency of the pulley system.

Machines

ICSE 2017

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Answer

Given,

V.R. = 4

L = 175 kgf

dL = 15 m

E = 50 kgf

(i) V.R. = dEdL\dfrac{dE}{dL}

Substituting the values we get,

4=dE15dE=15×4dE=60 m4 = \dfrac{dE}{15} \\[0.5em] \Rightarrow dE = 15 \times 4 \\[0.5em] \Rightarrow d_E = 60 \text{ m}

(ii) Work done by the effort = E x dE
     = 50 x 10 x 60
     = 3 x 104

(iii) M.A. = LE\dfrac{L}{E}

Substituting the values we get,

M.A.=17550=3.5\text {M.A.} = \dfrac{175}{50} = 3.5

(iv) Efficiency η = M.A.V.R.\dfrac{M.A.}{V.R.}

Substituting the values we get,

η=3.54=0.875η=87.5%η = \dfrac{3.5}{4} = 0.875 \\[0.5em] \therefore η = 87.5 \%

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