Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

The handle of a nutcracker is 16 cm long and anut is placed 2 cm from its hinge,if a force of 4 kgf is applied at the end of the handle to crack it, what weight, if simply, placed on the nut will crack it ?

Asked by Vikas 16th May 2016, 12:22 AM
Answered by Expert
Answer:

Hi,

Given that:

Distance of nut from fulcrum (d) = 2 cm

Distance of effort from fulcrum (D) = 16 cm

Load × Load arm = Effort × Effort arm

load × 2 cm = 4 kgf  × 16 cm

Load = 64 / 2 = 32 kgf

Answered by Expert 16th May 2016, 6:23 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!