Recent content by killaI9BI

Fapplied - Ffriction = Fnet - FgII where FgII is the gravitational force parallel to the incline ... according to my notes :wink:

that worked! I have learned about the FBDs but didn't think it'd really help. I will try them in the future. Thank you.

I'm still in left field somewhere

ohhh I see now; because it's moving at a constant velocity, the net force must be zero so the applied force equals the force of friction right?

I'm thinking the frictional force would be in the opposite direction as the applied force. I'm going to see if I can find another formula for net force that doesn't require the value of coefficient of friction.

Me either - just using equations from notes that I took during the lesson. I'm not really sure what it tells me about the net force... the formula that I have for net force is: Fnet = Fa – mg sin θ +or- μmg cos θ where Fa is applied force, m is mass, g is acceleration due to gravity...

Homework Statement You are pushing a 53 kg crate at a constant velocity up a ramp onto a truck. The ramp makes an angle of 22 degrees with the horizontal. If your applied force is 373 N, what is the coefficient of friction between the crate and the ramp? Homework Equations μ =...

lol thanks for your help guys!

:biggrin:yay!

I really do appreciate your input. I am pretty stunned when it comes to this stuff... I just don't understand how what I've done above is enough because as you can see here: Proving that equation is the last problem. I have no idea how to answer it.

just to be certain, can you confirm that what I've done looks right? thanks again!

I really appreciate your feedback. Thank you!

I don't have an excuse for my carelessness unfortunately :cry: I sincerely apologize for wasting your time. a) 7.5/5 = 1.5s 6/5 = 1.2s b) d = vi X t + 1/2a X t2 d = 1/2(-9.8) X 0.752 d = (-2.756) d = 2.8 m d = 1/2(-9.8) X 0.62 d = (-1.764) m d = 1.8 m My interpretation of...

face palm! d = vi X t + 1/2a X t2 d = ½(-9.8) X 1.12 d = (-5.929) m 12 – 5.929 = 6.071 m a) 6.1 m b) vi = (d/t) – ((a x t)/2) vi = (6.071/1.1) – (((-9.8) x 1.1)/2) vi = 5.519 – (-5.39) vi = 10.9 m/s Thank you very much!

the first 0.75 is half of the air time that I calculated from a). Sorry, I should've made that clear so you can follow what I'm trying to do :frown: I made the mistake of not changing my answers to positive values. I was calculating the height of the jump by considering the acceleration of...