Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
A bead slides without friction around a looptheloop. The bead is released from a height
20.3 m from the bottom of the looptheloop
which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?
 2 years ago
 2 years ago
A bead slides without friction around a looptheloop. The bead is released from a height 20.3 m from the bottom of the looptheloop which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?
 2 years ago
 2 years ago

This Question is Closed

krystal38garcia87Best ResponseYou've already chosen the best response.0
I already calculated speed. 11.112 m/s. To find the answer to this problem do I use the formula. F=mv^2/R = ((.005kg)(11.112 m/s)^2)/5m = 0.123N ?
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
A is the top of the loop?
 2 years ago

hoseinBest ResponseYou've already chosen the best response.1
if point "A" in top normal force is N=(mv^2/r)mg if "A" in bottom N equal to:dw:1329244817817:dw(mv^2/r)+mg
 2 years ago

hoseinBest ResponseYou've already chosen the best response.1
hence in bottom 2nd law of newton is Nmg=mv^2/r & on top is N+mg=mv^2/r
 2 years ago

krystal38garcia87Best ResponseYou've already chosen the best response.0
top... thainks guys.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.