Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
abdul_shabeer
Group Title
A man of weight W is standing on a lift which is moving upward with an acceleration a. The apparent wieght of the man is
 2 years ago
 2 years ago
abdul_shabeer Group Title
A man of weight W is standing on a lift which is moving upward with an acceleration a. The apparent wieght of the man is
 2 years ago
 2 years ago

This Question is Closed

Aadarsh Group TitleBest ResponseYou've already chosen the best response.2
See, I think the new weight would be: W1 = m(g + a) = mg + ma = W + ma = W + (W/g)a Please confirm with other experts also.
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
Yes your answer is right
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
Can you explain it in much more detail?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.2
Thanks Yaar.
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
How you got m(g+a)?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.2
see, when the object is moving upward, always add the acceleration due to gravity and the new acceleration. Multiply the sum by mass of the object. U will get the force acting on it.
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
The force acting on it would be the aapparent wieght of man?
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
apparent
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
Will it be the apparent weight of man?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.2
I think most probably it should be.
 2 years ago

abdul_shabeer Group TitleBest ResponseYou've already chosen the best response.0
Ok, Thanks a lot
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.2
U r welcome.
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.3
yes u have given the right formula but how? here it is when the lift is moving up, the normal force is graeter than the weight of man mg but apparent weight or weight in general of a body is determined by the NORMAL FORCE/NORMAL REACTION hence here Nmg=ma N=mg+ma again i repeat when we say the WEIGHT OF A BODY IS SOMETHING IT IS ACTUALLY THE NORMAL REACTION/FORCE AND NOT mg of the body
 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.