Question

A crowbar of length 45 cm is used to lift up a rock weighing 16 kg from the ground. If
the fulcrum is 5 cm from the rock, what is the effort needed (in kgf) to move the rock?​

Answer 1

Length of crow bar = 2m

length of load arm = 0.1M

length effort arm = length of crow bar -length of load arm

=2−0.1=1.9m

so, velocity ratio = length of effort arm/length of load arm = 1.9/0.1 = 19

use formula, velocity ratio = load/effort

19 = 100kgf/effort

effort = 100/19= 5.26 kgf

hence,5.26 effort is required.

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Answer 2

Answer:

length of crow bar =2m

length of load arm = 0.1 m

length of effort arm equals length of crowbar-length of load arm

=2-0.1=1.9m

Explanation:

so velocity ratio=length length of a effort

arm /length of load arm=

$\frac{1.9}{0.1} = 19$

velocity velocity ratio =load /effort

$19 = 100kgf \div effort$

$effort = \frac{100}{19} = 5.26kgf$

$hence \: 5.26 \: effort \: is \: reqired$