Posted by Jeanne on .
A box with a mass of 7.00 kg is at rest on a ramp that is at an angle of 20 degrees with the horizontal. The coefficient of static friction between the box and the ramp is 0.800. Use g = 9.80 m/s2. You now want to make the box move by applying a force. To start the box moving, what is the minimum force you need to apply if your force is directed ...
(a) parallel to the slope?
.N
(b) perpendicular to the slope?
.N
(c) horizontally?
N
I know that (a) is The static friction in the x direction minus the box sliding down in the x direction, but I cannot wrap my head around the other two.

Physics 
Anonymous,
i have this same problem but my coefficient of friction is .7 not .8. the answer i got for a) was 21.6614. i havnt gotten b or c. the equation i used for a. was F+Fgx'=fs

Physics 
Anonymous,
for b, the formula I used is (if P is the perpendicular force) Fgx'=uS(Fgy'P) (can you let me know if that is right. ive used 5 attempts before i figured out this formula and want to make sure. for , the formula i used was (if H is your horizontal force) Hx'+Fgx'=uS(Fgy'Hy') this one is correct

Physics 
Jeanne,
Can you explain what Hx'+Fgx'=uS(Fgy'Hy') means? It's hard to elaborate when the variables aren't defined.

Physics 
Jeanne,
same for Fgx'=uS(Fgy'P)

Physics 
Anonymous,
H is your horizontal force applied. it is your unknown Since the motion is on a ramp, you have to redefine your axis. the x' axis is parallel to the ramp and the y' axis is perpendicular to the ramp. so Hx' is the x' component of the H force. Hy' is the y' component of the H force. Fgx' is the x' component of gravity. uS is your coefficient of static friction.
does that make more sense? 
Physics 
Jeanne,
Yeah I got that, but I was hoping you could explain what you defined as the x' and y' component of each like in terms of cos and sin seeing that we might not be on the same page so i cant check ur answers exactly if i use something different.

Physics 
Anonymous,
also, P is the perpendicular force you apply. it is also your unknown. in the Hx'+Fgx'=uS(Fgy'Hy')Hx'=Hcos20 and Hy'=Hsin20

Physics 
Jeanne,
thus you assume Fgx' is mgcos20 and Fgy' is mgsin 20?

Physics 
Anonymous,
no, Fgy'=mgcos20 and Fgx'=mgsin20

Physics 
Anonymous,
ps, i just confirmed that if you plug in the correct numbers for those equations, you will get the right answers

Physics 
Jeanne,
okay, just making sure

Physics 
Jeanne,
I am trying your equation from for P. I have a feeling it will come out as a negative force and if you apply a perpendicular force it will only increase the contact force and thus not move the box, but let me see.

Physics 
Jeanne,
Yes, your equation is right so if you just plug in your given numbers for P=mgcos20(mgsin20/uS) you should get the correct answer.

Physics 
Anonymous,
yea, P is pushing the box away from the ramp not into it

Physics 
Jeanne,
yeahh