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

[sona1177]

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.

 

[ehild]

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

 

[sona1177]

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! :)