An effort of 50kgf is applied at the end of a lever of the second order, which supports a load of 750kgf, such that the load is at a distance of 0.1m from the hinge. find the length of the lever.

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Solution

1. Formula used:
Effort x Effort arm = Load x Load arm
2. Solution:
Given that,
Effort = 50kgf
load = 750 kgf
load arm = 0.1m
let effort arm = E
Effort x effort arm = Load x load arm
50 x effort arm = 750 x 0.1
effort arm = $\frac{75}{50}$ = 1.5m
3. Conclusion:
Hence, the length of the lever = 1.5m

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An effort of 50kgf is applied at the end of a lever of the second order, which supports a load of 750kgf, such that the load is at a distance of 0.1m from the hinge. find the length of the lever.

1. Formula used:Effort x Effort arm = Load x Load arm2. Solution:Given that, Effort = 50kgfload = 750 kgfload arm = 0.1mlet effort arm = E Effort x effort arm = Load x load arm50 x effort arm = 750 x 0.1effort arm = 7550 = 1.5m3. Conclusion:Hence, the length of the lever = 1.5m

1. Formula used:
Effort x Effort arm = Load x Load arm
2. Solution:
Given that,
Effort = 50kgf
load = 750 kgf
load arm = 0.1m
let effort arm = E
Effort x effort arm = Load x load arm
50 x effort arm = 750 x 0.1
effort arm = $\frac{75}{50}$ = 1.5m
3. Conclusion:
Hence, the length of the lever = 1.5m