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More mechanics!
A lift in a mine shaft takes exactly one minute to descend 500m. It starts from rest, accelerates uniformly for 12.5s to a constant speed which it maintains for some time and then decelrates uniformly to stop at the bottom of the shaft.
The mass of the lift is 5 tonnes and on the day in question it is carrying 12 miners whose average mass is 80kg
i) Sketch the speed time graph of the lift - done on attachment with some extra info i worked out from later questions.
During the first stage of the motion the tension in the cable is 53 640N.
ii) Find the acceleration of the lift during this stage = 0.8m/s2
iii) Fid the length of time for which the lft is travelling at constant speed =
40s, and find the final deceleration = 10/7.5 = 1.3333333333 m/s2
iv) What is the maximum value of the tension in the cable. I knwo the max value will be just before the lift comes to a stop to stop the downward deceleration the answer in the back of th book is 66.4kN.
I get this value if i use this T - 5960g = 5960*1.33333
T = 66354.67N but why is the reaction force of the miners 960g upwards ignored in the equation should it not be T + 960g -5960g = 5960*1.3333 in which case T = 56946.67. Do the miners feel this apparent weightlessness?
The mass of the lift is 5 tonnes and on the day in question it is carrying 12 miners whose average mass is 80kg
i) Sketch the speed time graph of the lift - done on attachment with some extra info i worked out from later questions.
During the first stage of the motion the tension in the cable is 53 640N.
ii) Find the acceleration of the lift during this stage = 0.8m/s2
iii) Fid the length of time for which the lft is travelling at constant speed =
40s, and find the final deceleration = 10/7.5 = 1.3333333333 m/s2
iv) What is the maximum value of the tension in the cable. I knwo the max value will be just before the lift comes to a stop to stop the downward deceleration the answer in the back of th book is 66.4kN.
I get this value if i use this T - 5960g = 5960*1.33333
T = 66354.67N but why is the reaction force of the miners 960g upwards ignored in the equation should it not be T + 960g -5960g = 5960*1.3333 in which case T = 56946.67. Do the miners feel this apparent weightlessness?
Related discussions
- BATH - Integrated Meng and Eeng
- Bristol (mech eng) or Bath (IMEE)
- Mechanical vs general engineering
- Mechanical vs Aerospace
- Bath engineering
- Mechanical Engineering Personal Statement
- material science and engineering vs mechanical engineering with foundation
- A levels for mechanics
- Electrical Eng at Bristol or Mechanical Eng at Sheffield
- Engineering fields in sustainability with a mechanical engineering degree??
- mechanical engineering or software engineering?
- Mechanical engineering or aerospace engineering
- Mechanical Engineering or Chemical Engineering?
- MSc in Advanced Mechanical Engineering vs MSc in Computing (Conversion Course)
- Integrated mechanical and electrical engineering vs pure mechanical
- Aerospace Engineerings vs. Mechatronics *NEED HELP*
- Mechanical Engineering Meng vs Aerospace Engineering Meng
- Need advice for applying to more than one type of engineering.
- General engineering vs Mechanical engineering
- Career from chemistry bsc or mechanical engineering BEng
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