What weight does the scale read when a man squats with an acceleration downward?

  • Thread starter Thread starter bem
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
When a man squats with a downward acceleration, the scale will read more than 80kg due to the increased normal force acting on him. The correct reasoning is based on the formula N = m(g + a), which indicates that the scale's reading increases with downward acceleration. A friend's suggestion that the scale would read less is incorrect. In extreme cases, such as free-falling while squatting, the scale would read zero. Confirming this understanding with a teacher or tutor is advisable for clarity.
bem
Messages
2
Reaction score
0
my pre-test has a question: "A man is standing on a balance scale which reads 80kg. Suddenly, he quats with a n acceleration a (downward). What weight does the scale read while the man is squatting?
(A)the same (B) more than 80kg (c) less than 80kg

I chosed (b)
because before we have N=mg
then after the man squatting N=m(g+a)
so the scale has to read more than 80kg. But then one of my friend said that N=m(g-a). I'm just confused. Can smbd help me ? I have a bet with him ( 10 bucks, hehe).
 
Physics news on Phys.org
Bem,

Consider the ultimate squat. What if the man squatted REALLY fast - i.e. he lifts his legs up off the scale and free falls. What would the scale read in that case?
 


Your reasoning is correct. When the man squats with an acceleration downward, the normal force acting on him will increase. This means that the scale will read a higher weight, which is option (b). Your friend may have made a mistake in their calculation or interpretation of the question. If you are unsure, you can always double check with a teacher or tutor to confirm your answer. Good luck with your bet!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top