Physics Problem Center Of mass HELP

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SUMMARY

The discussion focuses on calculating the acceleration vector for Jack, a skier with a mass of 85 kg skiing down a hill inclined at 35°. The correct acceleration vector, as stated in the textbook, is (3.95)i + (-2.29)j. The user initially calculated the acceleration as (3.83)i + (5.48)j, indicating a misunderstanding of the forces acting on Jack. Key issues identified include the misalignment of coordinate axes with the incline and the need to consider additional forces beyond gravity.

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FahimP
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Homework Statement


The problem is asking to find the Acceleration vector before jack reaches the less steep section.
the Ans from back of book says : (3.95)i + ( -2.29 ) j

My answer does not match at all : ANY HELP PLZZZZ

anyone know how to get this ?

Two skiers, Annie and Jack, start skiing from rest at different points on a hill at the same time. Jack, with mass 85 kg, skis from the top of the hill down a steeper section with an angle of inclination of 35°. Annie, with mass 70 kg, starts from a lower point and skis a less steep section, with an angle of inclination of 20°. The length of the steeper section is 175 m.



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Homework Equations





The Attempt at a Solution


This is how i tried to find the acceleration vector :

F = mg
F = 85kg( 9.81 ) = 568.98 N

568.98sin(35) = 326.35
568.98cos(35) = 466.05

a = Fnet/m
326.35/85 = 3.83
466.05/85 = 5.48

a= ( 3.83)i + ( 5.48 ) j
 
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I suspect that one problem is that the components into which you have resolved the gravitational acceleration vector g are components that are parallel and perpendicular to the incline. (Draw the vector decomposition and you'll see this is true). The problem is that your x and y coordinate axes are not aligned with these parallel and perpendicular directions (look at the figure).

Another problem is that Jack's weight is not the only force that acts on him (Hint: he never accelerates in the direction perpendicular to the incline, right? So something other than just gravity must be present). So, computing the net force in the x-direction and the net force in the y-direction requires a little bit more work and thought than what you've done here.

Draw a free body diagram for Jack.
 
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