Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

same for Fgx'=uS(Fgy'-P)

Answer 5

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?

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

no, Fgy'=mgcos20 and Fgx'=mgsin20

Answer 10

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

Answer 11

okay, just making sure

Answer 12

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.

Answer 13

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.

Answer 14

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

Answer 15

yeahh