Calculating Force Exerted on 17-kg Box from 90-N Force Acting on 11-kg Box

  • Thread starter Thread starter apples
  • Start date Start date
  • Tags Tags
    Confusing
AI Thread Summary
In a physics problem involving two boxes on a frictionless surface, a 90-N force is applied to an 11-kg box, which in turn exerts a force on a 17-kg box. The correct calculation shows that the acceleration for both boxes is 3.2 m/s², leading to a force of 35 N exerted by the 11-kg box on the 17-kg box. However, many participants mistakenly calculated the force as 55 N, citing Newton's third law, which states that forces between two objects are equal and opposite. Clarification indicates that the 11-kg box experiences a net force of 90 N minus the force exerted by the 17-kg box, while the 17-kg box only experiences the force from the 11-kg box. Understanding the dynamics and forces involved is crucial for solving such problems accurately.
apples
Messages
171
Reaction score
0

Homework Statement


Two boxes are sitting side on a frictionless surface. The box on the left has a mass of 11kg and the box on the right has a mass of 17kg. If a 90-N force pushes on the 11-kg box from the left, what is the force exerted on the 17-kg by the 11-kg box?

A) 55 N
B) 43 N
C) 68 N
D) 35 N

Homework Equations


F=ma

The Attempt at a Solution



F=MA
90=(11+17)a
a= 3.2 m/s^2
Since you have to find out the force exerted by the 11-kg box
F=11(3.2)
F=35N

This was a question on a quiz. All my friends got 55N.
 
Physics news on Phys.org
1...3.2m^s^2 is the correct acceleration for both boxes together
2...If you now draw afree body force diagram for both boxes you will see that the 11Kg box has two forces on it the resultant being 90-F where F is the force that the 17 Kg box puts on the 11Kg box
3...There is just one force on the 17Kg box ie F (Newtons 3rd law)
4...In your final step you chose to consider the 11Kg box but you should have wrote
90-F=11a.If you had chosen the 17Kg box you could write F=17a
 
Oops, I missed that last bit. Yes, 17x3.2 rather than 11x3.2.
 
Thanks for the reply
I don't understand what the force by the 17kg box on the 11kg box is.
 
The force wil be the same ie 55N. The left hand box pushes to the right with 55N and the right hand box pushes to the left with 55 N(3rd law )The result is that the left hand box has two forces :90 N pushing to the right and 55N pushing to the left making a resultant of 90N-55N.The right hand box has the single force of 55N pushing right.It would probably help if you sketched it out and marked in the forces.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top