Homework Help: Physics

Posted by Jeanne on Monday, October 10, 2011 at 11:28am.

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, Monday, October 10, 2011 at 2:08pm
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, Monday, October 10, 2011 at 4:22pm
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, Monday, October 10, 2011 at 4:55pm
Can you explain what Hx'+Fgx'=uS(Fgy'-Hy') means? It's hard to elaborate when the variables aren't defined.
Physics - Jeanne, Monday, October 10, 2011 at 4:57pm
same for Fgx'=uS(Fgy'-P)
Physics - Anonymous, Monday, October 10, 2011 at 5:00pm
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, Monday, October 10, 2011 at 5:01pm
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, Monday, October 10, 2011 at 5:02pm
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, Monday, October 10, 2011 at 5:03pm
thus you assume Fgx' is mgcos20 and Fgy' is mgsin 20?
Physics - Anonymous, Monday, October 10, 2011 at 5:04pm
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 - Anonymous, Monday, October 10, 2011 at 5:06pm
no, Fgy'=mgcos20 and Fgx'=mgsin20
Physics - Anonymous, Monday, October 10, 2011 at 5:10pm
ps, i just confirmed that if you plug in the correct numbers for those equations, you will get the right answers
Physics - Jeanne, Monday, October 10, 2011 at 5:13pm
okay, just making sure
Physics - Jeanne, Monday, October 10, 2011 at 5:14pm
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, Monday, October 10, 2011 at 5:19pm
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, Monday, October 10, 2011 at 5:24pm
yea, P is pushing the box away from the ramp not into it
Physics - Jeanne, Monday, October 10, 2011 at 5:26pm
yeahh

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