A steel shuttle is pulled along a rail, what is its initial acceleration?

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Homework Help Overview

The discussion revolves around determining the initial acceleration of a steel shuttle being pulled along a rail by a tension force. The shuttle has a mass of 760 grams and is subjected to a tension of 25N at a 45-degree angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the formula F/m = a to find acceleration, questioning the role of tension and the impact of friction on the system. There is also consideration of significant figures and the components of acceleration.

Discussion Status

Participants are actively exploring the problem, with some suggesting the need to account for friction in their calculations. There is recognition of potential errors in reasoning, but no consensus has been reached on the correct approach.

Contextual Notes

Friction is noted as a significant factor, with a coefficient of friction for steel on steel mentioned. The discussion also highlights the importance of significant figures in calculations.

potatotimer
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Homework Statement
Question: a machine has a 760g steel shuttle that is pulled along a square steel rail by an elastic cord. the shuttle is released when the cord has a 25N tension at a 45-degree angle what is initial acceleration?
Relevant Equations
F = ma or F/m = a
it sounds like you just need to use F/m = a which gives 25 / .76 = 32.89 = a
this didn't work so I tried to get the x component of acceleration so I did 32.89cos(45) = 23.26 and this didn't work.

what am I missing, the force is tension right so it should be 25N?
 
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potatotimer said:
Homework Statement:: Question: a machine has a 760g steel shuttle that is pulled along a square steel rail by an elastic cord. the shuttle is released when the cord has a 25N tension at a 45-degree angle what is initial acceleration?
Relevant Equations:: F = ma or F/m = a

it sounds like you just need to use F/m = a which gives 25 / .76 = 32.89 = a
this didn't work so I tried to get the x component of acceleration so I did 32.89cos(45) = 23.26 and this didn't work.

what am I missing, the force is tension right so it should be 25N?
Two possibilities come to mind. First, have you rounded to an appropriate number of significant figures? Second, have you been studying friction and, if so, would the friction of steel on steel come into play?
 
jbriggs444 said:
Two possibilities come to mind. First, have you rounded to an appropriate number of significant figures? Second, have you been studying friction and, if so, would the friction of steel on steel come into play?
yeah its friction
rereading made me realize that friction does need to be considered which for a steel on steel collision u = 0.8
so the acceleration is 25cos(45) - .8*.76*9.8
thanks for the help :)
 
potatotimer said:
- .8*.76*9.8
Two errors in there.
 

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