A block and tackle system have the velocity ratio $3$ . A man can exert a pull of $200 kgf$ . What is the maximum load he can raise with this pulley system if its efficiency is $50$ %? [ $5 marks$ ]

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Solution

Given:
velocity ratio $VR = 3$
Effort $E = 200 kgf$
Efficiency ( $η$ ) $=$ $50$ %
We know that when efficiency is $100 %$ then,
Effort ( $E$ ) = $\frac{Load}{n (number of pulleys)}$
$\Rightarrow 200 = \frac{L}{3}$

$\Rightarrow L$ $=$ $600 kgf$ [ $2.5 mark$ ]

When, efficiency is given as $50 %$ .

So, Maximum load that can be raised

= $\frac{50}{100} \times 600$ = $300 kgf$ [ $2.5 mark1$ ]

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A block and tackle system have the velocity ratio 3. A man can exert a pull of 200 kgf. What is the maximum load he can raise with this pulley system if its efficiency is 50%? [5 marks]

Given: velocity ratioVR=3 Effort E=200 kgf Efficiency (η) = 50% We know that when efficiency is 100% then, Effort (E) = Loadn (number of pulleys) ⇒200=L3 ⇒L= 600 kgf [2.5 mark] When, efficiency is given as 50%. So, Maximum load that can be raised = 50100×600 = 300 kgf [2.5 mark1]

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