An elevator starts from rest with a constant upward acceleration and moves 1m in the first 1.6s. A passenger is holding a 3.1kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8m/s^2.
What is the tension in the cord as the elevator accelerates? Answer in units of N.
2006-11-12 13:50:08 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The tension is the weight + the force Fb required to accelerate the bundle along with the rest of the elevator.
Need to calculate the elevator's acceleration a
y = (1/2)*a*t^2 where y = 1m and t = 1.6s. Solve for a.
Now calculate Fb
Fb = m*a. Solve for Fb.
T = W + Fb
2006-11-12 14:44:48 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
you should perceive all the forces in touch, and how they cancel one yet another out evaluate the rigidity of gravity, pulling the log immediately down, immediately in route of the middle of the earth, the burden rigidity Fw = ma Fw = (205kg)(9.8m/s^2) Fw = 2009 N The log is on a ramp, so this rigidity may have horizontal and vertical elements. On any ramp, the horizontal area of the burden rigidity may be found with Fwx = Fw(sin(attitude)), and the vertical ingredient is Fwy = Fw(cos(attitude)). Fwx = 2009N(sin(30)) = 1004.5 N Fwy = 2009N(cos(30)) = 1739.8 N utilizing the horizontal ingredient, which takes position to be perpindicular to the ramps floor, we may be able to make sure how a lot the ramp pushes decrease back; the conventional rigidity. Fn = 1739.8 N and then the rigidity of friction, given with Ff = Fn*cof of friction Ff = 1739.8N *.900 = 1565.9 N the rigidity of friction and the horizontal area of gravity are all the forces pulling the log DOWN the ramp. Ff + Fwx 1565 . 9 N + 1004.5 N = 2570.4 N the rope must have a rigidity more effective than 2570.4 newtons, with the intention to get the block transferring UP the ramp.
2016-11-29 02:13:51 · answer #2 · answered by Anonymous · 0⤊ 0⤋