piper210_355 Posted October 12, 2008 Author Posted October 12, 2008 ok so would the normal force still be mass times gravity in this problem or no.
big314mp Posted October 12, 2008 Posted October 12, 2008 No. Normal force is the force perpendicular to the plane. It is the green line (or the unlabeled blue, same thing) in the picture I posted. Mass times gravity is the red line. I'll list the relevant equations we have come up with, before I go to dinner: F(f) = u*F(n) F(f) = F(x) tan(theta) = F(x)/F(n) Rearrange those, and you should figure out theta. You don't actually need to find F(x), F(f), or F(n).
piper210_355 Posted October 12, 2008 Author Posted October 12, 2008 ok so I rearranged and I got this: tan(theta)=F(x)/F(n) tan(theta)=F(f)/[F(f)/u)] Here is where I'm not sure if i'm over simplifying(since in the other one i am dividing by a fraction i then change it by multipling by the reciprical) tan(theta)=F(f)*[u/F(f)] This cancels out the F(f) leaving... tan(theta)=u so then (theta)=tan^(-1) .45 (theta)=24.23 is that right? I really hope it is(I think it is). I'm pretty sure i know how to find the acceleration from here. All i need to know is if the angle is right.
piper210_355 Posted October 12, 2008 Author Posted October 12, 2008 :DYES:D Ok I got the rest of these steps from a video on youtube, but I just want to make sure it's right. First it said to get the force of gravity: F(g)=(gravity)(mass) F(g)=(9.81)(2.5) F(g)=24.5N Then I found the Force of the parallel (that's what they called the line in between F(n) and F(g) in the video). F(p)= F(g)sin(24.23) F(p)=(24.5)(sin 24.23) F(p)= 10.05 After that I used newton's second law to find the acceleration. a=F/m a=(10.05)/2.5 a=4.02m/s^2
big314mp Posted October 12, 2008 Posted October 12, 2008 Don't forget the coefficient of kinetic friction. You need to take that into account. Just calculate the force perpendicular (the normal force) using your trig, multiply by the coefficient of kinetic friction, and subtract from the force parallel. Then use that final force value for your last calculation. Other than that, excellent work!
piper210_355 Posted October 13, 2008 Author Posted October 13, 2008 ok so i'm going to put what i did just to make sure i understood/did it right tan 24.23=F(n) .45=F(n) F(n)F(k)=F(?what exactly am i calculating here?) (.45)(.28)=F(?) .126=F(?) F(p)-F(?)=F 10.05-.126=F 9.92=F a=F/m a=9.92/2.5 a=3.97m/s^2 I hope that's right thank you sooooooooooo much for your help
big314mp Posted October 13, 2008 Posted October 13, 2008 Use the same method you used to calculate the force parallel. F(perp) = F(g)*cos(24.23) The F(perp) is your normal force. You calculate the kinetic friction the same way you calculate the static friction. F(f) = F(perp)*u(kinetic) This force "takes away" from the force accelerating the box, so you subtract it from the force accelerating the box. In physics terms, this force is acting in an opposite direction from the force parallel. So subtract the kinetic friction force from the force parallel. The number that this gives you is the net force on the box. The net force, divided by the mass, will give you the acceleration.
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