What is the normal force on a child going down an inclined slide?

AI Thread Summary
A 23 kg child on a slide inclined at 38 degrees experiences forces including weight, normal force, and kinetic friction. The normal force is calculated using the equation Fnet = ma, leading to a determination that the normal force is approximately 177.6 N. It's important to correctly break down the weight into its x and y components, with the weight being mg = 225.4 N. The discussion emphasizes the need for precision in calculations and the importance of defining coordinate axes. Overall, the normal force calculation is confirmed to be correct in principle, with minor numerical adjustments suggested.
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Homework Statement


A 23 kg child goes down a straight slide inclined 38 degrees above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction. How large if the normal force of the slide on the child.


Homework Equations


Fnet=ma
Fnety=ay



The Attempt at a Solution



Fnet=ma
Fnety=may
Kfy + ny + wy=may=0 where Kf stands for kinetic friction, n stands for the normal force, and w stands for the weight.

0+ n-(222.4cos38)=0
0 + n-177.6=0
n=177.6

225.4 is the weight. I drew the triangle to find the x and y components of the weight such that the angle being given was = to the value of theta i used. Is 177.4 N for the the normal force correct? I want to make sure my equations are correct also because I just started doing problems where you have to break the weight into x and y components. Thanks.
 
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Well, it is correct in principle, but you should state the direction of your coordinate axis: the x-axis is parallel to the slope and points downward, y is normal to the slope and points upward.

Take a bit more care to the numbers you type in, and do not omit the units. The y component of the weight is mgcos(38°)=177.6 N, not 177.4, and mg=225.4 N, not 222.4.

ehild
 
ehild said:
Well, it is correct in principle, but you should state the direction of your coordinate axis: the x-axis is parallel to the slope and points downward, y is normal to the slope and points upward.

Take a bit more care to the numbers you type in, and do not omit the units. The y component of the weight is mgcos(38°)=177.6 N, not 177.4, and mg=225.4 N, not 222.4.

ehild

Thanks, I'll be more careful next time! :)
 

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