An elevator weighing 12 kN starts from rest and acquires an upward velocity of 2 m/sec in a distance of 5 m. If the acceleration is constant, what is the tension in the cable? *
We know that
Newton motion third
Law
"V^2=U^2+2aS"
"a=\\frac{V^2-U^2}{2S}"
Put value
"a=\\frac{2^2-0^2}{2\\times5}=0.4m\/sec^2"
Now We know that
"T-ma=mg"
"T=mg(1+\\frac{a}{g})=12000\\times(1+\\frac{0.4}{9.8})=12.489KN"
Comments
Leave a comment