Recent content by Pfloyd77

my actual shaft diameter is 2mm which would give me a gearbox ratio of of 4?

for my sums i made a new axis in my fbd that coincides with both N and F(friction) where the normal is equal to the force of fravity (mgsintheta) i don't understand how i have done my sums wrong as for the 2nd part i would need 1/1,772 m

the force i calculated was 17.72N i got this answer by using the following m=1.3kg g=9.8m/s^2 theta = 45 coefficient of static friction (us)= 0.9 N=mgsin45 F = (us)N +mgcos45 = (0.9)(1.3)(9,8)(sin45) +(1.3)(9.8)(cos45) F=17.71 since the motor can lift 0.01N at a 1m radius...

after I've calculated my net force how do i compare that with the force that is generated by my motor...what formula(s) would i need to use

how do go i go about doing step 2 using all of my motor values we never covered any motor equations in class so I am really new at the motor equations and knowing how to apply them

I have to build a monorail that can climb a 45 degree slope length of the track is 8m. The mass of the monorail is .3kg and it has to carry a mass of 1kg up the slope.im given a 15V electric dc motor with: Power = 6W at motor stall no load rpm at 0.07 Amp = 3800 stall torque at 0.4 amp = 0.01N...