How Do Forces Act on an Elevator in Motion and at Rest?

AI Thread Summary
Forces acting on an elevator at rest include gravitational force (F_g) downward and tension (F_T) upward, which are equal at 29,400 N. When ascending at a constant speed of 3.0 m/s, the net force is zero, meaning tension equals gravitational force. If descending with an acceleration of 3.0 m/s², the tension is calculated to be 20,400 N, while gravitational force remains at 29,400 N. The discussion emphasizes the importance of a consistent sign convention and understanding net forces in relation to acceleration. Properly analyzing these forces is crucial for solving related physics problems.
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A stationary elevator and its contents have a combined mass of 3000Kg. The elevator is suspended by a single cable. (Assume three significant digits.)
Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest.
If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point?
If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point?

For a i did---F_g=mg = 3000*9.8=29400N (Down)
F_g=〖-F〗_T
F_T=29400 N (Up)

For c i did---Net Force downwards=mg-T
downward acceleration= Net force/m= (mg-T)/m
a = g-T/m
T= (g-a)m
= (9.8-3)3000= 20400N(Up)
Net Force acting upwards= mg= 3000*9.8= 29400N(Down)

Is this right? How to do part b?what equation should i use?
 
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It's best to stick with a sign convention, so let's say positive is up and negative is down, ok? So that would make;

Fg = -mg = -29400 ; T = -Fg = 29400

So, now we have the forces at rest. Now, the net force at all times can be written as Fnet = T+Fg. So you have your basic equations set up, all you have to do it equate them with Newton's Second law.
 
But i part b, they have given the speed, not the acceleration. How do use that in Ma=T+F_g?
 
pinkyjoshi65 said:
But i part b, they have given the speed, not the acceleration. How do use that in Ma=T+F_g?
Indeed, if an object is traveling at constant speed, what is it's acceleration? What does this mean the net force acting on the object is?
 
if the elevator is traveling at a constant speed then the a is 0. That means the T and F_g will be equal, so the net force will b 0. Right?
 
pinkyjoshi65 said:
if the elevator is traveling at a constant speed then the a is 0. That means the T and F_g will be equal, so the net force will b 0. Right?
Spot on. And for (c)...?
 
for c: net force= T+F_g
and we know a=f/m...so a= T+F_g/m
hence then T will be equal to m(a-g)= -20400N. hence tension will be 20400N down, and F_g will be 29400 N Up. Right?
 
pinkyjoshi65 said:
for c: net force= T+F_g
and we know a=f/m...so a= T+F_g/m
hence then T will be equal to m(a-g)= -20400N. hence tension will be 20400N down, and F_g will be 29400 N Up. Right?
Careful, your numbers are right but check your sign. How do you propose the tension acts downwards and gravity acts upwards?

You were correct in your first post, I just wanted to make sure you understood the importance of defining a definite coordinate system. :smile:

As an aside, an easier method would be just to consider the force required to accelerate the elevator at 3 m/s^2. :smile:
 
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Thank you so much...:)
 
  • #10
i had another question..
A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
B) what is the tension in teh string at the bottom of the swing?
C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?
Here is what i could do:
I drew a vertical circle with 2 positions a and b. A is at the top and B is at the bottom. so at A the F_g is down and T is up. at B,T is down and F_g is up. Right? then what?
 
  • #11
pinkyjoshi65 said:
i had another question..
A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
B) what is the tension in teh string at the bottom of the swing?
C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?
Here is what i could do:
I drew a vertical circle with 2 positions a and b. A is at the top and B is at the bottom. so at A the F_g is down and T is up. at B,T is down and F_g is up. Right? then what?
:bugeye:
 
  • #12
i dint understand..so tension is down for A? and T is up for B and F_g is down for B?..k then how do i proceed?
 
  • #13
pinkyjoshi65 said:
i dint understand..so tension is down for A? and T is up for B and F_g is down for B?..k then how do i proceed?
Two points to remember;

(1) The tension always acts towards the centre of the string.

(2) Gravity always acts towards the centre of the earth.

Next, I would draw a FBD for each case individually.
 
  • #14
so then at A net force will b mg-T, and at b net force will b T-mg. right? but mg has to b 0 since for part A, weight is 0. that means net force at A will b equal to -T. and we knoe tht centripital force=Mv^2/r. So that will be -T=-mg=0.2*v^2/0.75. when i solve this i get v=2.71m/s. is that right?
 
  • #15
pinkyjoshi65 said:
so then at A net force will b mg-T?
No, at A both forces act in the same direction, i.e. down. Like I said previously, the tension in a string always acts towards the centre of the string.
 
  • #16
but then the net force at A will still be -T.so the v will still b 2.71m/s
 
  • #17
pinkyjoshi65 said:
but then the net force at A will still be -T.so the v will still b 2.71m/s
Nope I'm afraid it isn't. The question asked for the tension when the speed is zero...
 
  • #18
no we have to find the speed when the weight is 0. dosent that mean that we hve to find the speed when mg=o.that means that only tension force is actin at A. isint that right..?
 
  • #19
pinkyjoshi65 said:
no we have to find the speed when the weight is 0. dosent that mean that we hve to find the speed when mg=o.that means that only tension force is actin at A. isint that right..?
:redface: My bad, I misread the question, you are indeed correct, your solution is right.
 
  • #20
so at B, the tension will b 1.96 N?
 
  • #21
pinkyjoshi65 said:
so at B, the tension will b 1.96 N?
That would be the case if the velocity is zero, but v > 0 at the bottom.
 
  • #22
how do i find the T at B then?I know that at B the net force will be T-mg. oh wait do i use a as 9.8m/sq sec at B and use that to find the net force. Then i can find T. is that right?
 
  • #23
pinkyjoshi65 said:
how do i find the T at B then?I know that at B the net force will be T-mg. oh wait do i use a as 9.8m/sq sec at B and use that to find the net force. Then i can find T. is that right?
Think about energy conservation. When your rock was at the top, you had some non-zero potential energy. When your rock reaches the bottom of the circular path, where's this potential energy gone?
 
  • #24
changes into kinetic energy since it is moving..?
 
  • #25
pinkyjoshi65 said:
changes into kinetic energy since it is moving..?
Yup, so now you can work out how fast it's moving at the bottom and hence the required centripetal force, thus leading you to the tension.
 
  • #26
but wht is the value for KE..we don't know tht..
 
  • #27
pinkyjoshi65 said:
but wht is the value for KE..we don't know tht..
You can use conservation of energy to work it out...

(1) What is the change in height between the top and the bottom of the path?

(2) What is the total energy at the top of the path?

(3) All the energy in (2) is converted to kinetic at the bottom of the path, so what's the velocity at the bottom of the path?
 
  • #28
change in h will be 0.75m, total energy at the top of the path will b the potential energy=mgh=1.47J. so at B the KE will be 1.47J since no PE exits. right?
 
  • #29
tht give me T=5.87N. What about part C?
 
  • #30
pinkyjoshi65 said:
change in h will be 0.75m, total energy at the top of the path will b the potential energy=mgh=1.47J. so at B the KE will be 1.47J since no PE exits. right?
You might want to check that change in height... draw a picture...
 
  • #31
The h frm the centre to point A is 0.75m and the h frm the centre to point B is also 0.75. so the total distance between A and B is 1.5m. change in height will be 1.5?..
 
  • #32
pinkyjoshi65 said:
The h frm the centre to point A is 0.75m and the h frm the centre to point B is also 0.75. so the total distance between A and B is 1.5m. change in height will be 1.5?..
Sounds good to me :approve:
 
  • #33
sp doing that i got v as 5.42m/s and F will b 7.83 N. and substituting in T-mg we get T as 9.79N. what about part C? how do i find the angle
 
  • #34
pinkyjoshi65 said:
sp doing that i got v as 5.42m/s and F will b 7.83 N. and substituting in T-mg we get T as 9.79N. what about part C? how do i find the angle
Well, suppose the centre of the path remains the same, how does the energy of the rock change?
 
  • #35
it continuously changes into kinetic energy since its in motion..?
 
  • #36
pinkyjoshi65 said:
it continuously changes into kinetic energy since its in motion..?
It will have some kinetic energy, but it will also have some potential energy since it is some height above the base of the verticle path. Do you follow?
 
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  • #37
oh yes...!..now i get it..so how can i use it to find the angle?
 
  • #38
pinkyjoshi65 said:
oh yes...!..now i get it..so how can i use it to find the angle?
The trouble is, the angle will depend on the velocity, which will depend on the height above our zero potential, which depends on the angle. If you haven't done so I suggest you draw yourself a diagram. The string will form the hyp on a right-angled triangle, with the base being \sin\theta (where \theta is the angle between the string and the horizontal) and the vertical side being \cos\theta. Also mark on the point B of our vertical circle (this is our zero potential) and let h be the vertical height of our rock above the point B.

Have you got the picture straight?
 
  • #39
not really..you lost me..can you explain it again..
 
  • #40
pinkyjoshi65 said:
not really..you lost me..can you explain it again..
Sorry, let me try and clarify;

  1. Draw a right triangle
  2. lable the hypotenues R, this is your string.
  3. lable the vertical side \cos\theta
  4. lable the horizontal side \sin\theta
  5. lable the angle between the hypotentues and the vertical side \theta
  6. Draw a circle at the corner where the hypotenues meets the horizontal side, this is your rock.
  7. Extend the verticle line down some arbitrary distance and draw a circle at the end of your line
  8. Label the distance between the cicle you just drew and the corner between the hypotentues and the verticle side h. This is your height above the base.

Is this a better picture?
 
  • #41
yes this is much better..then..?
 
  • #42
pinkyjoshi65 said:
yes this is much better..then..?
Draw the forces that are acting on your rock.
 
  • #43
ok..so there r 2 forces acting tension (up) and mg down..then..
 
  • #44
pinkyjoshi65 said:
ok..so there r 2 forces acting tension (up) and mg down..then..
The tension isn't actually acting upwards, it will be acting along the string, but a component of the tension will be acting upwards. Do you follow?
 
  • #45
No..
 
  • #46
pinkyjoshi65 said:
No..
It's difficult to explain without the aid of a diagram, does this <http://www.ngsir.netfirms.com/englishhtm/ConicalPendulum.htm> make anymore sense? If it doesn't try googling 'conical pendulum'.
 
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  • #47
ok now i got it..so the diagram is like this: the hypotunese is the radius, the base is sin(theta), the vertical line is cos(theta). the rock is attached to the hypotunese. 2 forces act on the rock, mg: downwards and T along the hypotunese. Theta is between the hypotunese and the vertical line(costheta)..then..cos theta + plus a certain dist is costheta+h
 
  • #48
So, what can you say about the vertical component of the tension when compared with the force of gravity acting on the rock?
 
  • #49
the vertical component of the tension is acting in the oppostie direction to the mg of the rock..?
 
  • #50
looks as if hootenanny's gone, i'll have a quick bash,

the opposing forces are the same, as grvaity is the only force acting on the rock. At any set time the vertical componet of the tension of the string = mg.

When you know the vertical component of the tension you can then find the hypotenuse, which should be the tension in the string,

sound right to u?
 

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