An elevator starts from rest with a constant
upward acceleration andmoves 1m in the first
1.9 s. A passenger in the elevator is holding a
7.4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the ele-
vator accelerates? The acceleration of gravity
is 9.8 m/s2 .
1
Expert's answer
2012-02-09T08:25:07-0500
Using the Newton's Second Law the tension can be found from the equation: Tension = F = ma, where a is the sum of the acceleration of gravity and the acceleration of elevator: a = g + aElev. As the elevator moves with a constant acceleration, let's write an equation for displacement: S = V0*t + (aElev*t^2)/2. As the initial speed V0 = 0, then: S = (aElev*t^2)/2. So, aElev = 2S/t^2. Consecuently, We get the formula for tension: T = m(g + 2S/t^2) = 7.4*(9.8+2*1/1.9^2) = 76.6 N.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment