Active forum topics
- Hydraulics: Rotating Vessel
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
New forum topics
- Hydraulics: Rotating Vessel
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
Recent comments
- Determine the least depth…1 week 2 days ago
- Solve mo ang h manually…3 weeks ago
- Paano kinuha yung height na…3 weeks ago
- It's the unit conversion…1 month ago
- Refer to the figure below…3 weeks 6 days ago
- Yes.4 months 2 weeks ago
- Sir what if we want to find…4 months 2 weeks ago
- Hello po! Question lang po…5 months 1 week ago
- 400000=120[14π(D2−10000)]
(…6 months 1 week ago - Use integration by parts for…7 months 1 week ago
Re: HELP PO!
Let us call the required pressure Normal Force
letter (a)
Normal Force = mg = 80(9.81) = 784.8 N.
This is also true when the elevator is running at constant velocity (zero acceleration). The pressure he exerted on the floor will change when the acceleration of the elevator is not zero.
letter (b)
Weight = 784.8 N downward
Normal Force = 784.8 - 80 = 704.8 N
letter (c)
Weight = 784.8 N downward
Normal Force = 784.8 + 80 = 864.8 N
Conclusion: Exercise may reduce the weight by 10 kg in less than a month. :)