Question

In a test on machine an effort of 50N was required to raise a load pf 250N. If the effort moves through 150 mm to raise the load by 25 mm, find the following:

i. The Mechanical Advantage

ii. The Velocity Ratio

iii. The efficiency

iv. The work done by the effort.

v. The work done in raising the load 25 mm.

vi. The power developed by the effort over a period of 5 seconds

Answer 1

Solution.

$F_e=50N;$

$F_l=250N;$

$d_e=150mm=0.150m;$

$d_l=25mm=0.025m;$

$i. MA=\dfrac{F_l}{F_e};$ $MA=\dfrac{250N}{50N}=5;$

$ii. VR=\dfrac{d_e}{d_l};$ $VR= \dfrac{0.150m}{0.025m}=6;$

$iii. \eta=\dfrac{ F_ld_l}{F_ed_e}= =\dfrac{250N\sdot0.025m}{50N\sdot0.150m}=0.83;$

$iv. A_e=F_ed_e=50N\sdot0.150m=7.5J;$

$v.$ $A_l=F_ld_l=250N\sdot0.025m=6.25J;$

$vi. N=\dfrac{A_e}{t}=\dfrac{7.5J}{5s}=1.5W;$

Answer: $i. 5;$

$ii. 6$ ;

$iii.0.83;$

$iv. 7.5J;$

$v. 6.25J;$

$vi. 1.5W;$