Finding accleration and tension as an elevator moves upwards/downwards.

AI Thread Summary
The discussion centers around calculating the acceleration and tension in an elevator using the forces acting on it. For part A, the acceleration is determined to be -2.79 m/s² when the scale reads 424 N, indicating the elevator is accelerating downward. In part B, with a scale reading of 639 N, the acceleration is found to be 0.76 m/s², indicating upward movement. The tension in the cable is calculated to be 3034 N for part A and 1259 N for part B, with the tension increasing when the elevator accelerates upward. The calculations and understanding of the forces involved are confirmed to be correct, emphasizing the relationship between acceleration direction and cable tension.
Crusaderking1
Messages
158
Reaction score
0

Homework Statement

A 593 N physics student stands on a bathroom scale in an 875 kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 424 N.

A. Find the magnitude of the acceleration of the elevator.

B. What is the acceleration if the scale reads 639 N?

C. What is the tension in the cable in part A?

D. What is the tension in the cable in part B?

Homework Equations



ƩF = ma

T= ma +mg

The Attempt at a Solution



A. 424-593 =60.5a
-2.79 = a

B. 639-593= 60.5a
0.76 = a

C. (875)(2.79)+(593) =T
3034 N = T

D. (875)(.76) + 593
1259 N = T

Is this right? Thanks.

Also, would acceleration stay negative for part C?
 
Physics news on Phys.org
Lets deal with this question first.
"Also, would acceleration stay negative for part C?"

Well what did you calculate as the acceleration in part a)?
Why would you think that answer would change?

and then parts c) and D)
Which way is the elevator accelerating in part A) and Part B) ie up or down.

If the elevator is accelerating up or down, in which case then would the tension in the cable be greater or lessor than before the elevator started moving?

Draw a FBD to obtain all the forces on the elevator.

And what does this mean in the problem
" 875 kg (including the student) elevator "
 
256bits said:
Lets deal with this question first.
"Also, would acceleration stay negative for part C?"

Well what did you calculate as the acceleration in part a)?
Why would you think that answer would change?

and then parts c) and D)
Which way is the elevator accelerating in part A) and Part B) ie up or down.

If the elevator is accelerating up or down, in which case then would the tension in the cable be greater or lessor than before the elevator started moving?

Draw a FBD to obtain all the forces on the elevator.

And what does this mean in the problem
" 875 kg (including the student) elevator "

You're right, part A is negative so I'm guessing I use that same value for part C.

The elevator is accelerating down in part A and up in part B.

Tension would be greater as it accelerates upward.

I will try to re-solve for C and D and answer back, but I think part A and B are right. Thanks.
 
Last edited:
ok

I think upwards, T-mg=ma
So, T-8575=665
T= 9240 N

I think downwards, T-mg=-ma
So, T-8575=-2441.25
T= 6134 N

Upwards a=0.760
Downwards a=2.79

Sound better?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top