Equilibrium applications of Newton's law of motion II

In summary, the normal force exerted by the table on box 1 is equal to the combined weight of boxes 1 and 2, which is 80.7N. The tension in the rope and the weight of box 3 do not affect the normal force.
  • #1
pookisantoki
44
0
Box 1 is resting on the table with box 2 resting on top of box 1. a massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box and the other end is connected to box 3. The weights of the three boxes are w1=55.0N, W2=25.7N. Determine the mangitude of the normal force that the table exerts on box 1.

So i drew it out and the tension and normal force would both go up and the weight would go the opposite direction...
So would I add (M*g)+tension? then subtract the weight? where would box three come into play??
Thank you
 
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  • #2
I don't think the tension has anything to do with this... Unless I'm misunderstanding this problem, box 1 is flat on a table with 2 on top of 1, and box 3 is suspended in the air by a rope-pulley system. So the weight of box 3 would create the tension in the rope for the (box1/box2) weight. However, in this problem, the tension is parallel to the table, whereas we are looking for the normal force which is always perpendicular to the surface. So my best guess would be that the normal would just be (mass of box1+mass box2)g since the box is being held still on the table. If I'm misunderstanding the problem, let me know.
 
  • #3
So i did what you said and (55+38.8)9.8 and got 919.24 but it was wrong so I'm assuming that box 3 would create upward tension?? since it is hanging??
 
  • #4
well the thing is that the weight of the box is a downward force and it is canceled out by the tension in the rope which is an upward force, so that should be in equilibrium assuming that the box is not in motion
 
  • #5
maybe I'm still misunderstanding, is there a way you could attach a picture of the system?
 
  • #6
Ahhh i think I see where your problem is, you added the WEIGHTS together and multiplied by g, you needed to add the MASSES together and multiply by g. Luckily this problem is easier than you thought, all you need to do is add the two weights together since weight=mass*g. That should give you your normal force.
 

Related to Equilibrium applications of Newton's law of motion II

1. What is the concept of equilibrium in terms of Newton's law of motion?

Equilibrium in terms of Newton's law of motion refers to a state where the forces acting on an object are balanced, resulting in a constant velocity or no movement at all. This means that the net force on the object is zero, according to Newton's first law of motion.

2. How is equilibrium related to Newton's second law of motion?

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In equilibrium, the net force is zero, so the acceleration is also zero, meaning that the object remains at rest or moves with a constant velocity.

3. What are some real-life applications of equilibrium and Newton's second law of motion?

Equilibrium and Newton's second law of motion have many practical applications, including calculating the weight and tension in a bridge, analyzing the forces acting on a car during a turn, and determining the center of mass in objects like buildings and airplanes.

4. Can equilibrium be achieved in all situations?

No, equilibrium can only be achieved in situations where the net force on an object is zero. If there is an unbalanced force acting on an object, it will accelerate in the direction of the force until a new equilibrium is reached.

5. How can we use Newton's second law of motion to solve equilibrium problems?

To solve equilibrium problems, we can use Newton's second law of motion to set up and solve equations for the forces acting on an object. By setting the net force equal to zero, we can determine the unknown forces or variables involved in the situation.

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