- #1
Jacobim
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Text of the problem:
Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, whereas a point at the poles experiences no centripetal acceleration. Assume the Earth is a uniform sphere and take g = 9.800 m/s2.
(a) If a person at the equator has a mass of 68.0 kg, calculate the gravitational force (true weight) on the person. (Give your answer to four significant figures.)
(b) Calculate the normal force (apparent weight) on the person. (Give your answer to four significant figures.)
I have already used the tutor application on this webassign and solved the problem, it just doesn't make sense to me. Here is the tutor text:
At the equator, the centripetal acceleration, ac is downward. Taking as positive the direction away from the center of the Earth, we have
n − mg = −mac.
Solving for the normal force from this equation, gives the person's apparent weight as
n = mg − mac
= m(g − ac)I do not understand why it should not be n = mg + mac
It seems to me the magnitude of the normal force should equal the sum of the forces caused by gravitational acceleration and centripetal acceleration.
usually n=mg so the net Fy is zero if on a flat surface. so adding in force from acceleration it should simply be
n=mg+mac, and then the person would feel heavier not lighter.
Thanks for any help!
Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, whereas a point at the poles experiences no centripetal acceleration. Assume the Earth is a uniform sphere and take g = 9.800 m/s2.
(a) If a person at the equator has a mass of 68.0 kg, calculate the gravitational force (true weight) on the person. (Give your answer to four significant figures.)
(b) Calculate the normal force (apparent weight) on the person. (Give your answer to four significant figures.)
I have already used the tutor application on this webassign and solved the problem, it just doesn't make sense to me. Here is the tutor text:
At the equator, the centripetal acceleration, ac is downward. Taking as positive the direction away from the center of the Earth, we have
n − mg = −mac.
Solving for the normal force from this equation, gives the person's apparent weight as
n = mg − mac
= m(g − ac)I do not understand why it should not be n = mg + mac
It seems to me the magnitude of the normal force should equal the sum of the forces caused by gravitational acceleration and centripetal acceleration.
usually n=mg so the net Fy is zero if on a flat surface. so adding in force from acceleration it should simply be
n=mg+mac, and then the person would feel heavier not lighter.
Thanks for any help!
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