Apparent Weight. A 610 N physics student stands on a bathroom scalein an 850-kg

Question

Apparent Weight. A 610 N physics student stands on a bathroom scalein an 850-kg (including the student) elevator that is supported bya cable. As the elevator starts moving, the scale reads510 N. (a) Find the acceleration of the elevator(magnitude and direction -- take up to be the positivedirection).
m/s2

(b) What is the acceleration if the scale reads 730 N?
m/s2

(c) If the scale reads zero, what is the acceleration of theelevator?
m/s2


(d) What is the tension in the cable in parts (a) and (c)? cable tension in part (a) N cable tension in part (c) N (a) Find the acceleration of the elevator(magnitude and direction -- take up to be the positivedirection).
m/s2

(b) What is the acceleration if the scale reads 730 N?
m/s2

(c) If the scale reads zero, what is the acceleration of theelevator?
m/s2


(d) What is the tension in the cable in parts (a) and (c)? cable tension in part (a) N cable tension in part (c) N cable tension in part (a) N cable tension in part (c) N

Explanation / Answer

if a student stands on a bathroom scale with a 610 N reading thenwith F=mg we have 610=m(9.8), m=62.2449 kg and now F=m(g+a)    with F=510 N, g=9.8, and m=62.2449 kg wehave 510=62.2449(9.8+a) 520=610+62.2449a a=-1.45 m/s^2 (b) so now F=m(g+a)   where F=730 N, g=9.8, m=62.2449 kg 730=62.2449(9.8+a) 730=610+62.449a a=1.93 m/s^2 (c) so now F=m(g+a) where f=0 N, g=9.8, m=62.2449 kg 0=62.2449(9.8+a) 0=610+62.449a a=9.77 m/s^2 (d) from Newton's second law we have that Fnet=ma however we know that Fnet equals the sum of all theforces acting on the elevator, thus Fnet=T+W with T=tension and W=weight, settingFnet=Fnet gives us ma=T+W or T=ma-W so for part a. T=(850)(-1.45)-Melev.g Melev.=(850 kg-62.2449)=787.755 kg T=850(-1.45)-(787.755)(9.8) T=-8952.5 N and for part c. T=850(9.77)-(787.755)(9.8) T=584.501 N

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