Recent content by kelvin scott

ok thank you ,.

ok so (10*1)+(50*3)+(20*8)=R2*6 10+150+160=r2*6 320 =R2*6 320/6 =R2= 53.33 R1*6=(10*5)+((50*3) - (20*2) R1*6 =50+150-40 R1 =160/6 = 26.67 my attempt

ok so the moment sum should be R1 2R1 = (10*1) +(50*6)+(20*8) 2R1= 10 + 300 +160 R1 = 470/2 R1 =235 and R2 8R2 = (10*5) +(50*8) +(20*8) 8R2 =50 +400+160 R2 = 610/8 R2 = 76.25 not sure I am right on the length, and sorry for the delay

ok

Homework Statement a beam is 10 m long with point loads of 10kN at 3m from left and 20kN 10m from left. R1 and R2 these supports are 2m in from either end of the beam. The beam has a UDL of 5kN/m-1 along its length. Homework Equations (a) determine the vertical reaction at the support(b)...

ok thanks to all who contributed

yea thanks but can't locate the keys to achieve this t2 0.5 = 0*1.5+0.5*a*1.5 ^2 0.5 =0.5*a*1.5^2 =a*1.5^2/ 0.5 =a*1.5^2=1 a =1/ 1.5^2 a =1/2.25 a =0.44m/s^2 hoping I am on the right track

so it means ; d= VoT + 0.5 at (2) ?

A mass of 0.5kg is suspended from a flywheel is released from rest and falls a distance of 0.5m in 1.5s ,calculate the linear acceleration of the mass Σ Torque = I x a Σ F = m* a Im I right with the formulas?

I want to be very good with motors and control systems