Alright so we got a 40cm (diameter) 35kg wheel on an incline (22 degree). There is a force pulling the wheel from the top of the wheel. Find a) the required force F (i‘m guessing this means so that the wheel doesn‘t roll down the slope) and b) the minimum coefficient of friction required between the barrel and slope. im thinking for the first part you do moments about the bottom of the wheel (contact section) and find F (components for the moment would be from the force and CG). And then in part B you use that F and do a normal fiction/incline problem. Is this the correct thinking? Cheers
The gravitational force down the incline is Fg m*g*sinΘ The forces up the incline are the pulling force F + the frictional force Ff ?*m*g*cosΘ By symmetry, F Ff, and by Newton's 3rd law, 2F Fg Ergo, a) F ?Fg ?*m*g*sinΘ 64.245 N Since Ff F, ?min*m*g*cosΘ ?*m*g*sinΘ → ?min ?tanΘ b) ?min ?tan22° .202
I think you have the right idea a) a safe assumption is that the wheel is symmetrical so the force exerted at the top of the wheel (tangentially) would be equal to the force exerted by the bottom of the wheel on the ramp so Fp Fw sin(angle) m g sin22 b) minimum mu would leave Fp Ff Fp mu Fn mu m g cos22 solve for mu
The torque caused by the forces causes the wheel to roll. I choose the pivot point at the point where the wheel touches the incline surface. The component of the weight that is parallel to the incline is causing the wheel to accelerate down the incline. This force is located at the center of the wheel, which is 20 cm from the incline. Force parallel mass * g * sin θ 35 * 9.8 * sin 22 128.5 N The torque caused by this force 35 * 9.8 * sin 22 * 0.20 25.698 N*m This torque causes the wheel to roll down the inline. The force pulling the wheel from the top of the wheel is causing the wheel to accelerate up the incline. This force is located at the top of the wheel, which is 40 cm from the incline. Torque F * 0.40 This torque causes the wheel to roll up the incline. The friction force is located at the surface, so, the friction force produces no torque. Torque caused by force on top of wheel Torque caused by force at center of wheel F * 0.40 25.698 F 64.245 N Force parallel down the incline – Force up the incline 128.5 – 64.245 62.245 N (down the incline) This is the total force is directed down the incline, so the friction force must be directed up the incline. Ff ? * mass * g * cos θ ? * 35 * 9.8 * cos 22 ? * 35 * 9.8 * cos 22 64.245 ? 0.202 Sorry about my math error!
