Finding the Correct Force for Acceleration in Kinetics of Translation

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SUMMARY

The discussion centers on calculating the force required to accelerate a 200kg body supported by wheels and a sled with a coefficient of friction (uk) of 0.5. The user initially calculated the force (P) to be 491.78N, while their professor indicated the correct value should be approximately 2350N. Key equations used include Efx=max, Efy=0, and Emcg=0, along with the frictional force equation Fa=uNa. The discrepancy in the solution suggests a need to re-evaluate the equations applied, particularly the first and third equations.

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Homework Statement


A 200kg body is supported by wheels @ B to roll freely w/o friction & by a sled @ A with which uk = 0.5. Compute the value of P to cause an acceleration of 0.02g. (g=9.8m/s^2)

NOTE: Image not drawn to scale.
http://www.geocities.com/timothybridge/kinOftrans.JPG

Homework Equations



Efx=max

Efy=0

Emcg=0

Fa=uNa

The Attempt at a Solution



P-Fa=200(0.02)(9.8)

Na+Nb=1960

Nb(1.5)-Na(1.5)+P(1.2)-Fa(1.8)=0

Fa=0.5Na

solving 4 equations simultaneously,

P=491.78N
Na=905N
Nb=1055N

My professor said the answer should be around P=2350N

Can someone identify what's wrong with my solution?
 
Last edited:
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Recheck your third and first equation.
 

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