Calculating the normal force with Newton's Laws

In summary, the acceleration of block A is due to the pushing force of 49.6 N and the additional force of normal force between A and B.
  • #1
abswors
4
0
Homework Statement
Hey I could really use some help with this question. I know I need to use Newton's Laws but I am not sure if the force I am calculating when I us ethe equation F=ma is the right one. Please help:

Block A (of mass: mA = 500g) and block B (mB = 3kg) are placed in contact with each other on a frictionless surface. A force of 49.6 N is applied to block A pushing it into block B resulting in a normal force between the blocks. Calculate the strength of the normal force between block A and block B.

I would appreciate any help!
Relevant Equations
f=m*a
Here's what I have:
Block A is accelerating at a= 49.6/0.5 m/s (99.2m/s)
Block B force is F= 3kg * 99.2 = 297.6N
but I'm not sure where to go from here?
 
Physics news on Phys.org
  • #2
First of all, acceleration is measured in m/s^2, not in m/s. Getting units correct is very important.

Second, your first equation describes the acceleration of block A if the total force on A was 49.6 N. Is the pushing force the only force acting on A? Have you drawn free body diagrams for A and B?
 
  • #3
Yes I have, sorry about the units error I wrote that very quickly. The pushing force is the only force acting on A in this system and I do have a free body diagrams. I am just not sure what information to use to find the normal force. I have attached the full question in case I left something out.
 

Attachments

  • Screen Shot 2020-04-23 at 6.35.40 pm.png
    Screen Shot 2020-04-23 at 6.35.40 pm.png
    19.9 KB · Views: 139
  • #4
abswors said:
The pushing force is the only force acting on A in this system
Then why does B move?
 
  • #5
haruspex said:
Then why does B move?
It moves because the two blocks are in contact, so there is a force.
 
  • #6
abswors said:
It moves because the two blocks are in contact, so there is a force.
Yes, A must be exerting a force on B, so what forces are acting on A?
 
  • #7
abswors said:
Yes I have
No, you have not. You have shown a helpful graphic of the situation but that is not a FBD. Do you understand what a FBD is? You need to show all the forces in play.
 
  • #8
haruspex said:
Yes, A must be exerting a force on B, so what forces are acting on A?
That is what I am trying to calculate: the normal force between A and B
 
  • #9
abswors said:
That is what I am trying to calculate: the normal force between A and B
Ok, so we are agreed there are two forces acting on A, not just one as you wrote in post #3, yes?

Remember that you are free to choose subsystems. In the present problem, it will be convenient to start with the subsystem consisting of the pair of blocks. This allows you to ignore the normal force between them because it is internal to the subsystem. So now you have just one external force acting on the combined mass.

What does that tell you about the acceleration of this subsystem?
 
  • #10
Your "key mistake" as already pointed out but not emphasized is that the only force acting on A is the pushing force of 49.6N. There is one more force acting on A. Which one is it?
Once you take into account this force, then by applying Newton's 2nd law once for block A and once for block B you ll get a system of two equations with two unknowns: the normal force and the common acceleration of the blocks.
You will also need Newton's 3rd law to handle a small detail regarding the normal force (from block A to block B and from block B to block A).
 
  • #11
You can calculate the acceleration of the combined A and B resulting from the total force applied. Then calculate how much of that force is required to accelerate B that much. That will be the force between A and B.

PS. If this helps you to get the solution, be sure that you understand how it can be methodically shown in a free body diagram.
 

Related to Calculating the normal force with Newton's Laws

1. What is the normal force?

The normal force is a contact force that acts perpendicular to the surface of an object when it is in contact with another object or surface. It is also known as the support force, as it is responsible for supporting an object's weight.

2. How is the normal force calculated using Newton's Laws?

The normal force can be calculated using Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In the case of calculating the normal force, the acceleration is equal to zero as the object is at rest. Therefore, the normal force is equal to the object's weight, which is calculated by multiplying its mass by the acceleration due to gravity.

3. Can the normal force be greater than the weight of an object?

Yes, the normal force can be greater than the weight of an object. This can happen when an object is on an inclined plane or when there are other forces acting on the object, such as a person pushing down on it. In these cases, the normal force will be equal to the sum of the weight of the object and the additional forces acting on it.

4. Is the normal force always equal to the weight of an object?

No, the normal force is not always equal to the weight of an object. As mentioned before, the normal force can be greater than the weight of an object in certain situations. It can also be less than the weight of an object, such as when an object is in free fall or when it is accelerating downwards.

5. How does the normal force affect an object's motion?

The normal force does not directly affect an object's motion, as it is a contact force and acts perpendicular to the direction of motion. However, the normal force can indirectly affect an object's motion by balancing out other forces, such as gravity, and keeping the object in equilibrium. Without the normal force, an object would not be able to stay in place or move in a specific direction.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
261
  • Introductory Physics Homework Help
Replies
7
Views
377
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
972
  • Introductory Physics Homework Help
Replies
13
Views
552
  • Introductory Physics Homework Help
Replies
9
Views
901
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top