Calculating Initial Acceleration for a Steel Object on a Rail | 21 N Tension

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SUMMARY

The discussion focuses on calculating the initial acceleration of a steel object weighing 0.89 kg, pulled by a cord with a tension of 21 N at a 45-degree angle. The participant correctly applies Newton's second law (F=ma) and separates forces into x and y directions, ultimately deriving the acceleration equation. Key friction coefficients are provided: 0.8 for static friction, 0.6 for kinetic friction, and 0.002 for rolling resistance. The importance of checking whether static friction is overcome before proceeding with calculations is emphasized, along with the necessity of using significant figures in measurements.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of friction coefficients (static, kinetic, rolling resistance)
  • Ability to resolve forces into components (x and y directions)
  • Familiarity with trigonometric functions (sin and cos)
NEXT STEPS
  • Learn about the implications of static vs. kinetic friction in motion analysis
  • Study the concept of rolling resistance and its distinction from static friction
  • Explore the calculation of normal force in inclined scenarios
  • Investigate the role of moment of inertia in rolling objects
USEFUL FOR

Students in physics or engineering courses, educators teaching dynamics, and anyone involved in mechanics or motion analysis of objects on inclined planes.

rysezhae
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Homework Statement



I need to find the initial acceleration of a steel object that weighs .89 kg, that is pulled along a steel a rail by a cord. The cord is at an 45 degree angle and has a tension of 21 N. The friction coefficents(spelled right?) of steel on steel are: .8 for static friction, .6 for kinetic friction and .002 for for rolling friction.

Homework Equations


F=ma
Fp=21 N


The Attempt at a Solution


What i did was separate the net forces into the x direction and the y direction. In the x direction i got Fpcos(theta)-Fk=max (x = x direction). I then solved for ax giving me (Fpcos(theta)-Fk)/m.
Since Fk =ukFn (kinetic friction coefficient * normal force), i substituted that in for Fk in ax. In the y direction i got, Fpsin(theta)+Fn-Fe=0. I solved for Fn then substituted that back in for Fn in the ax equation getting: (Fpcos(theta)-uk(Fe-Fpsin(theta)))/m. No sig figs required, since the problem doesn't say anything about it
 
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##\cos(45^\circ)=\sin(45^\circ)=1/\sqrt{2}##

I'd have wanted to see you check that static friction is overcome first.

You require sig figs whenever you have measurements in your calculations - or do you think the values given are infinitely accurate?

Did you have a question?
 
rysezhae said:
.002 for rolling friction.
What's rolling here? If the object can roll then you might need to worry about moment of inertia. (I dislike the term 'rolling friction'. Sometimes people confuse it with static friction. It is often not a form of friction, but rather of imperfect elasticity. 'Rolling resistance' is better.)
 

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