Recent content by MPat

Homework Statement A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown. a) Draw a free body diagram for the beam, clearly labelling all the forces acting on it b) Calculate the tension in the cable...

Ohhh...I think I see your point. If my starting point was B then the rod would have already gained some KE from A to B. So if I use conservation of energy and start at position A, the height would be 1.0m. mgh (point A) = KE (point C).

Whoops! You're totally right. I lost a g in there. L/2*mg =1/3ml^2α Solve for α m cancels out 3Lg/2L^2 = α L cancels out (3g)/(2L)=α 3(9.8)/2(1)= α = 14.7 rad/s^2Found the h or change in h by taking half of the length of the rod. L=1.0m, H = L/2 If the rod is starting at position B with rod...

Homework Statement Consider a uniform rod of mass 12kg and length 1.0m. At it's end the rod is attached to a fixed, friction free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine, a) the angular acceleration of...

Thank you! I think I got it, it's essentially the hypotenuse formed by the triangle...my numbers weren't adding up because I was forgetting that I had to separate the vectors into components, N-S and E-W.

Homework Statement A 15 000kg loader traveling east at 20km/h turns south and travels at 25km/h. Calculate the change in the loader's a) kinetic energy b) linear momentum Homework Equations ΔKE= 1/2mvf^2 –1/2mvi^2 Δp= mv2 –mv1 m= 15000 vi = 20km/h = 5.56m/s vf= 25km/h = 6.94m/s The Attempt at...

GOT IT!...answer is Us = 0.22

Thanks for your help! Much appreciated.

Ok that correcting that I get the following sinθ (sinθFfr/cosθ + mg/cosθ) + cosθFfr = mv^2/r =sinθtanθFfr + tanθmg + cosθFfr = mv^2/r = Ffr(sinθtanθ + cosθ) + tanθmg = mv^2/r Ffr = (mv^2/r - tanθmg)/(sinθtanθ+cosθ) Even if I isolate Ffr here, m is unknown and I haven't been able to cancel it out...

Under vertical component? Vertical component cosθFn=mg Fn=mg/cosθ Eq 2 I divide both sides by cosθ here on the left side cosθ cancels out and on the right it becomes mg/cosθ Is this what you were referring to?

Homework Statement If a curve with a radius of 88m is perfectly banked for a car traveling at 75km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/hr 75km/h = 20.8m/s 95km/h = 26.4m/s Homework Equations Fr=mv^2/r Fr=centripetal force m = mass r...

Okay... g=v^2/r r = 6371 km = 6371000m v=sqrt g*r v= 7901m/s v=2πr/T T=2πr/v T=2π*6371000/7901 T=5066 seconds T= 84.4 mins So a day on Earth should be 84.4 mins so that an object can float freely above the equator. One thing I'm still confused about: Fr=mv^2/r Fr=mg - I came to that...

Ahh I'm confused. So if I take v=gT/2π and solve for T i get T=2πr/v Eq 2...what do I do with that? I don't have any of the values to help me solve for T. You said g=v^2/r Eq. 1 was correct...My assumption would have been that Fr would have to equal 0 in order for the object to float freely...

So for the object to float Fr should equal 0 correct?

Homework Statement Earth is a spherical object, which completes a full rotation every 24 hours. How long should the day on Earth be so that an object at the equator is able to float freely above the ground? Homework Equations v=2πr/T Fr=mv^2/r v=velocity r= radius T = time for comleting 1...