Question

A crowbar of length 45 cm is used to lift a rock weighing 16kg 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

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} = 190.11.9=19

velocity velocity ratio =load /effort

19 = 100kgf \div effort19=100kgf÷effort

effort = \frac{100}{19} = 5.26kgfeffort=19100=5.26kgf

hence \: 5.26 \: effort \: is \: reqiredhence5.26effortisreqired

Answer 2

Length of crow bar = 2m

length of load arm = 0.1M

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

=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.

hope it helps (ʘᴗʘ✿)

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