Instrument attached to wire - newton's laws


by cdlegendary
Tags: attached, instrument, laws, newton, wire
cdlegendary
cdlegendary is offline
#1
Feb2-10, 07:23 PM
P: 16
1. The problem statement, all variables and given/known data

A 5.10 kg instrument is hanging by a vertical wire inside a space ship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 290 m in 10.0 s with constant acceleration.

a.) Draw a free-body diagram for the instrument during this time.(Assume that the space ship is accelerating upward. )

b.) Find the force that the wire exerts on the instrument.

2. Relevant equations

f=ma
v=d/t
a=v/t

3. The attempt at a solution

I already drew the FBD, and it's just 2 vectors, the tension and weight, tension upward with more magnitude, and weight downward with less magnitude than that of tension.

The part I'm having trouble is with part 2, shouldn't the tension just be the weight of the instrument attached to the wire + the force of acceleration due to the ship?

So, T = mg + ma

T = (9.8m/s^2)(5.1kg) + (2.9m/s^2)(5.1kg)

This seems to be wrong though. (I got the 2.9m/s^2 through the v = d/t and a = v/t equations.)
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Spinnor
Spinnor is offline
#2
Feb2-10, 07:42 PM
P: 1,362
Quote Quote by cdlegendary View Post
1. The problem statement, all variables and given/known data

A 5.10 kg instrument is hanging by a vertical wire inside a space ship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 290 m in 10.0 s with constant acceleration.

a.) Draw a free-body diagram for the instrument during this time.(Assume that the space ship is accelerating upward. )

b.) Find the force that the wire exerts on the instrument.

2. Relevant equations

f=ma
v=d/t
a=v/t

3. The attempt at a solution

I already drew the FBD, and it's just 2 vectors, the tension and weight, tension upward with more magnitude, and weight downward with less magnitude than that of tension.

The part I'm having trouble is with part 2, shouldn't the tension just be the weight of the instrument attached to the wire + the force of acceleration due to the ship?

So, T = mg + ma

T = (9.8m/s^2)(5.1kg) + (2.9m/s^2)(5.1kg)

This seems to be wrong though. (I got the 2.9m/s^2 through the v = d/t and a = v/t equations.)
v = d/t is average velocity so this is not a relevant equation.

But distance, d = a*t^2/2 for constant acceleration.

I got a = 2.4 m/s^2 not the 2.9 you got.
cdlegendary
cdlegendary is offline
#3
Feb2-10, 07:47 PM
P: 16
Quote Quote by Spinnor View Post
v = d/t is average velocity so this is not a relevant equation.

But distance, d = a*t^2/2 for constant acceleration.

I got a = 2.4 m/s^2 not the 2.9 you got.
Ah I see. I thought I could use the average velocity to get the acceleration. I guess not. Well, when I use the d = (at^2)/2 equation, I get the acceleration to be 5.8m/s^2, not 2.4. What am I doing wrong?

a = 2d/t^2 = 2(290)/(10^2)

Spinnor
Spinnor is offline
#4
Feb3-10, 06:06 AM
P: 1,362

Instrument attached to wire - newton's laws


Quote Quote by cdlegendary View Post
Ah I see. I thought I could use the average velocity to get the acceleration. I guess not. Well, when I use the d = (at^2)/2 equation, I get the acceleration to be 5.8m/s^2, not 2.4. What am I doing wrong?

a = 2d/t^2 = 2(290)/(10^2)
My mistake.


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