Solve the following numerical problems. a) A load of 400N is lifted up by an effort of 100N. If load distance is 20cm, what will be the effort distance? (Ans: 80cm) b) Two boys, Shrijan having weight 600N and Shrijesh having weight 300N are playing see-saw. If Shrijan is sitting at 2m from fulcrum, where should Shrijesh sit from fulcrum to balance Shrijan?(Ans: 4m) c) A lever of length 1m has been used ttoko lift a load of 600N by applying an effort of 200N. If load is at 20cm from fulcrum, calculate mechanical advantage, velocity ratio and efficiency. (Ans: MA = 3, VR = 4, n=75%) d) Study the figure below and find the value of effort. (Ans: 120N) Muyn) 2.5m 600N 0.5m ? Science and Enyin​

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Explanation:

given,

load = 400 N

effort = 100 N

load distance = 20 cm

we know that ,

E*Ed = L*Ld

=100 N* Ed = 400N * 20 cm

=100N * Ed = 8000N/cm

= Ed =( 8000N/cm ) / 100N

= Ed = 80 cm

b. soln.

given,

load = 600 N

load distance = 2 m

effort = 300 N

effort distance = ?

we know that ,

= E *Ed = L * Ld

= 300N * Ed = 600N * 2 m

= 300N * Ed = 1200N/m

=Ed =( 1200N/m ) / 300 N

= Ed = 4 m

C. soln.

given,

load = 600 N

load distance =20 cm

effort = 200 N

effort distance = ?

M.A = ?

V.R = ?

Efficiency = ?

we know that ,

= E *Ed = L *Ld

= 200N * Ed = 600 N * 20 cm

=200 N *Ed = 12000 N/cm

=Ed = ( 12000 N/cm) / 200 N

= Ed = 60 cm

Also,

M.A = load / effort

=600 N / 200 N

= 3

V.R = Ed/ Ld

= 60 cm / 20 cm

= 4

efficiency = ( M.A / V.R ) 100 %

= ( 3 / 4 ) 100%

= 75 %

d. soln.

given,

load = 600 N

load distance = 0.5 m

effort distance = 2.5 m

effort = ?

we know that ,

= E * Ed = L * Ld

= E * 2.5 m = 600 N * 0.5 m

= E * 2.5 m = 300 N / m

= E = ( 300 N / m ) / 2.5 m

= E = 120 N

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explanation:

it is easily magnetized.