Page 100 - CITS - WCS - Mechanical

WORKSHOP SCIENCE - CITS

Effort x Effort arm = Load x Load arm
17 (x) = 100 x 50
100 x 50 17 X =
x =100x50\17= 294.1 cm
= 294.1 cm

: Distance between the load and point of force
= 294.1-50
= 244.1 cm
= 2.4410 m Ans

2 A simple screw jack having pitch = 1 cm and the effective length of the lever = 50 cm. What shall be the velocity
ratio? A force of 5 kg is applied on lever and lifts a load of 1100 kg. What is its efficiency?
Solution

Length of level / =50 cm
Pitch of screw p = 1 cm
Load lifted = 1100kg
Effort applied = 5kgs

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CITS : WCS - Mechanical - Exercise 8 CITS : WCS - Mechanical - Exercise 8