1.) An elevator (mass 4700 kg ) is to be designed so that the maximum accelerati
ID: 1656152 • Letter: 1
Question
1.) An elevator (mass 4700 kg ) is to be designed so that the maximum acceleration is 0.0620 g . Part A What is the maximum force the motor should exert on the supporting cable?
Part B
What is the minimum force the motor should exert on the supporting cable?
2.) Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated seats.
Part A
For a simple 16 upward ramp, what minimum length would be needed for a runaway truck traveling 170 km/h ?
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
upload force gives maximum force
Apply Newton second law
F nett = may
T- mg = ma
T- 4700(9.8) = (4700) (0.0620(9.8)
T = 48915.72 N
down ward force gives minimum force
mg- T = may
(4700)(9.8) - T =(4700)(0.0620*9.8)
T = 43204.28 N
(2)
the net force along x direction
Fx= - mg sin thets
a= g sin theta
Apply kinematic equation
x- xo = v^2 - vo^2/ 2a
= 0- vo^2/ 2 ( - g sin theta)
= vo^2/ 2 g sin theta
= ( 170 ( 1000 m/ 3600 s)62 / 2 ( 9.8 sin 16)
=412.76 m
rounding to two significant digits
s= 410 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.