When Should 'g' Be Negative in Physics Equations?

  • Thread starter Thread starter jteh
  • Start date Start date
  • Tags Tags
    Negative
AI Thread Summary
In physics equations, the sign of 'g' depends on the chosen coordinate system and the direction of motion. When defining a coordinate system, if the direction of gravity aligns with the positive axis, 'g' is positive; if it opposes the positive direction, 'g' is negative. For force calculations using F=mg, 'g' should always be positive to maintain clarity in force direction. Textbooks often assume upward as positive when dealing with upward motion, while downward motion may be treated as positive if the object is falling. In uncertain situations, it is advisable to take upward as positive and use -9.81 m/s² for downward acceleration.
jteh
Messages
2
Reaction score
0
Hi PF, just a quick question that always seems to confuse me.. when should 'g' be negative in equations? Intuitively, I tend to always substitute -9.81 m/s into equations but it seems that this can be incorrect in special cases.

Are there certain situations where 'g' should be or shouldn't be negative??

Thank you!
Jess.
 
Physics news on Phys.org
For kinematics equations, when you're using g as the acceleration, define a coordinate system first. That is, define an axis and choose a direction to be positive. If the acceleration due to gravity is in that direction, you use positive g. If the acceleration is in the opposite direction, you use negative g.

For equations like F=mg, you never want to use -g in place of g. This is because you want F to be positive, since positive forces are easier to think about.
 
Thank you so much! Great explanation.
 
Just adding to ideasrule's excellent explanation...

I've noticed that textbooks tend to use upward as the positive direction (a=-9.81 m/s2) when the motion is upward, or at least is upward at the beginning of a projectile's trajectory. Downward might be taken as positive (a=+9.81 m/s2), but only when the motion is always downward, i.e. the object is initially falling or it is dropped from rest.

If you're ever in doubt in a kinematics problem, just take upward as positive and use -9.81 m/s2.

p.s. Welcome to Physics Forums, jteh.
 
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