Why Does the Desk Lamp Move Despite Equal and Opposite Forces?

  • Thread starter Thread starter Mo
  • Start date Start date
  • Tags Tags
    Law
AI Thread Summary
Newton's third law explains that when one body exerts a force on another, the second body exerts an equal and opposite force back. When pushing a desk lamp, the force from the hand does not cancel out because they act on different bodies, allowing the lamp to move. The movement is determined by the ratio of force to mass, which results in acceleration for the lamp. The hand and body experience a much smaller acceleration due to their larger mass compared to the lamp, and friction prevents noticeable movement of the body. Understanding these interactions clarifies why forces occur in pairs and how they affect motion.
Mo
Messages
81
Reaction score
0
Hello all, need a little bit of help understanding Newton's third Law.

It states that:

" While body A exerts a force on body B, body B exerts an equal and opposite force on body A"

Okay, i can understand that.I also understand that the force are acting upon different bodies.

so (and here's the question..) if i push my desklamp across the .. desk.. then i am exerting a force in one direction on the lamp, and it is exerting an equal and opposite force on my hand.

If so, why can i still get the desklamp to move. In my textbook it tells me that the forces do not cancel each other out, because they act on different bodies, but .. still ... i obviously don't understand it all that well.

Okay the next question from the book was

Explain why forces occur only in pairs"

Is it simply enough to say: Whenever one body exerts a force on a second body, the second body always eerts a force on the first body, hence forces occur only in pairs.

Help + Advice please!

Regards,
Mo
 
Physics news on Phys.org
Mo said:
Hello all, need a little bit of help understanding Newton's third Law.

It states that:

" While body A exerts a force on body B, body B exerts an equal and opposite force on body A"

Okay, i can understand that.I also understand that the force are acting upon different bodies.

so (and here's the question..) if i push my desklamp across the .. desk.. then i am exerting a force in one direction on the lamp, and it is exerting an equal and opposite force on my hand.

If so, why can i still get the desklamp to move. In my textbook it tells me that the forces do not cancel each other out, because they act on different bodies, but .. still ... i obviously don't understand it all that well.

The 3 laws of Newton do not act independent of each other.U'll have to use them at the same time.The answer to your 'dilemma' is that,according to the second law,to each force corresponds an acceleration.In your case,the hand that pushes the lamp gives it an acceleration equal to the ratio between the force and the lamps mass.The total force acting on the lamp is just the force u're using to push it along the table.Of course,nthere's the friction force,too,but that,if the coefficent is small,doesn't alter the results significantly.

Daniel.
 
Thanks for your reply, but I am still not 100% , am i right in saying that the reason that my hand never acclerated off in the opposite direction (after all the lamp is exerting a force on my hand in the opposite direction) is because of the ratio between force and mass (making it a very small acceleration?)
 
Mo said:
Thanks for your reply, but I am still not 100% , am i right in saying that the reason that my hand never acclerated off in the opposite direction (after all the lamp is exerting a force on my hand in the opposite direction) is because of the ratio between force and mass (making it a very small acceleration?)

It's not only your hand,yer whole body should be accelerating in the opposite direction with a tiny acceleration (due to the big ratio of masses:yours & the lamps).You don't move,becuase of the friction between your feet and and the floor,or between your butt and the chair your butt uses to relax...

Daniel.
 
Thanks for your help :)

Regards.
Mo
 
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