Acceleration & Normal Force of a Box: 4 m/s2 & 418 N

AI Thread Summary
The discussion revolves around calculating the acceleration and normal force acting on a box, with the paper stating the values as 4 m/s² and 418 N, respectively. Participants clarify that the normal force is not always equal to the weight of the box and emphasize the importance of understanding the net forces acting on the box. A misunderstanding about the forces involved led to an incorrect calculation of acceleration, initially resulting in 8 m/s² instead of the correct 4 m/s². The final resolution highlights that the box is on a horizontal surface, and the correct approach involves balancing the forces to find the normal force and acceleration. Ultimately, the calculations were corrected, confirming the values stated in the paper.
Remle
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Homework Statement
A 25 kg box is being pushed down on by an applied force of 200 N inclined at 60° below the horizon.
Relevant Equations
Normal force: FN=m×g
  1. What is the acceleration of the box? Paper says the answer is 4 m/s2.
  2. What is the Normal force acting on the box? Paper says the answer is 418 N.
I know that for most cases FN=Fg=W. So, by definition the "original" Normal force is 245.25 N (am I correct?)
I calculated the Fay which is -173.205081 N, so I think that I have to add this two? Then the "new" FN and Fg would be 418 N.

Also when I do my calcs for acceleration I end up with 8 m/s2 not 4.

This is what I have:
Screenshot 2023-03-03 102212.png
 
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This box must be on some kind of surface to have a normal force exerted on it. Is the surface horizontal or inclined. What does the problem say? Your answers will depend on that clarification.

What you call the "original" normal force is the weight. So call it that to avoid confusion.
Remle said:
I calculated the Fay which is -173.205081 N, so I think that I have to add this two?
Which two? Also, please limit your numbers no more than 3 significant figures.
 
This box must be on some kind of surface to have a normal force exerted on it. Is the surface horizontal or inclined. What does the problem say? Your answers will depend on that clarification.
The problem does not say anything about a surface, so I just assumed it because that's how a normal force works.
What you call the "original" normal force is the weight. So call it that to avoid confusion.
When I say original is I calculate FN with mass and gravity. Then I start adding or substracting other forces (if there's any) to this FN.
Which two? Also, please limit your numbers no more than 3 significant figures.
Fg=245.25N and Fay=-173.205 N
 
Remle said:
I end up with 8 m/s2 not 4.
How ?
 
BvU said:
How ?
$$F_\rm{a_\rm{y}} = -200 \times \sin(60) = -173.205~N$$
$$F_\rm{a_\rm{x}} = 200 \times \cos(60) = 100~N$$
$$F_\rm{g}= 25 \times -9.81=-245.25~N$$
$$F_\rm{Net_\rm{y}} = - 245.25 - 173.205 + 245.25 = -173.205~N$$
$$F_\rm{Net_\rm{x}} = 100~N$$
$$F_\rm{Net}=\sqrt{{(-173.205)^2} + 100^2} = 200.00~N$$
$$a = \frac{200} {25} = 8~\rm{m/s^2}$$
 
Remle said:
$$F_\rm{Net_\rm{y}} = - 245.25 - 173.205 + 245.25 = -173.205~N$$
The normal force doesn't automatically equal the weight of the box. It is a force acting on the box, which you must solve for.
 
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Remle said:
$$F_\rm{Net}=\sqrt{{(-173.205)^2} + 100^2} = 200.00~N$$

You are reconstructing the given 200 N o_O ! Do you think the acceleration is in that direction ?

[edit] PS: kudos for the typesetting !

##\ ##
 
Remle said:
I know that for most cases FN=Fg=W
Confuses you. Not useful.

The ##F_\text{net}## concept also confuses you. Suggestion: memorize Newton in the form $$\vec F_\text{net} = m\vec a$$

In your exercise there is a box on a smooth horizontal surface. Without the 200 N extra force the net force is zero: the weight force from gravity is balanced by the normal force. So ##\vec a = 0##.

With the 200 N extra force , you correctly look at its components. And write ##F_\text{net, y}= F_{\text g} + F_\text{a, y} + F_\text{N}\ .\ ## Since there is no acceleration in the vertical direction, we must have ##F_\text{net, y}=0##. Which allows you to calculate the normal force.

In the horizontal direction, you (also correctly) write ##F_\text{net, x}= F_\text{a, x} \ . \ ## There are no other horizontal forces (no friction), so ##m\,a_x = F_\text{a, x} \ .\ ## Which allows you to calculate the acceleration.

That's all there is to it :wink: !

##\ ##
 
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BvU said:
Confuses you. Not useful.

The ##F_\text{net}## concept also confuses you. Suggestion: memorize Newton in the form $$\vec F_\text{net} = m\vec a$$

In your exercise there is a box on a smooth horizontal surface. Without the 200 N extra force the net force is zero: the weight force from gravity is balanced by the normal force. So ##\vec a = 0##.

With the 200 N extra force , you correctly look at its components. And write ##F_\text{net, y}= F_{\text g} + F_\text{a, y} + F_\text{N}\ .\ ## Since there is no acceleration in the vertical direction, we must have ##F_\text{net, y}=0##. Which allows you to calculate the normal force.

In the horizontal direction, you (also correctly) write ##F_\text{net, x}= F_\text{a, x} \ . \ ## There are no other horizontal forces (no friction), so ##m\,a_x = F_\text{a, x} \ .\ ## Which allows you to calculate the acceleration.

That's all there is to it :wink: !

##\ ##
You are reconstructing the given 200 N o_O !
OMG... feel dumb. The ## 200~N ## is already the diagonal force. So the only thing I have to do is find ## F_\rm{ay} ## and it add to ## F_\rm{g} ##. Then only force left is ## F_\rm{ax} ## which that gives me ## 100~N ## dividing that with ## 25~\rm{kg} ## provides the ## 4~\rm{m/s^2} ##.

Thank you BvU. Just need another mind to help me see things differently.
 
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BvU said:
[edit] PS: kudos for the typesetting !

##\ ##
@Remle ,

:welcome:

And I also say "kudos for the typesetting!"
 
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