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