A person stands on a scale on the ball of one foot at an angle of 13.3
a) Force= F=mgcos=67.5(10)cos13.3
" mgsin
will be the
component of gravity whcin is balanced by
=656.8N frictional force."
b) F= 67.5(10)cos30=584N The force is changed by 72.8N
c) The normal force on one foot
will be equal to = mgcos=656.8N for part
a) and 584N for part b)
For two feet Newtons second law of motion could be written
mgcos-Fn-Fn=m(0)
mgcos=2Fn
Fn= mgcos/ 2
Fn= 686.8/2
=343.4N
This will be the normal force when he is standing on both feets.
In case of 30 degree angle put the value of cos30 in above formula.
A person stands on a scale on the ball of one foot at an angle of...
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