Box pushing another box Dynamics

In summary, Sammy S found that the minimum acceleration needed to keep block m2 from sliding down the face of block m1 is 1915N. However, without some clarification, it's not possible to determine Fpush without knowing if the coefficients of friction also apply to the horizontal surface and the bottom of box m1.
  • #1
yolo123
63
0
Hi, I have an exam soon. This is urgent. Can you check my answer?

A box with mass 100kg is being pushed with enough force such that a second box with mass 2kg does not slide down and the two accelerate to the right together.
If us=0.5 and uk=0.1, what is acceleration minimum such that box m2 does not slide (please see picture.)? What is minimum value of Fpush?

What would change if block m1 is moving at constant speed?

I find that Fpush minimal is 1915N.

I find that acceleration minimal is 17.8 m/s^2.

If block moves at constant speed, the block would also move at constant speed in x direction but be affected in y direction because of gravity and just fall.
 

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  • #2
Please I know this is lots of work, but you've got to help me.
 
  • #3
why don't you post your solution instead of just posting your answer?
 
  • #4
dauto said:
why don't you post your solution instead of just posting your answer?

Exactly. If you post your calculations, we can seek for mistakes in them. Now the best we can do is to redo everything from scratch.
 
  • #5
I think that 1915N is wrong.
[tex](1915N-0.1 \cdot 100kg \cdot g ) \cdot \frac{2}{102} \cdot 0.5 = 2kg \cdot g[/tex]
[tex]1915N-10kg \cdot g = 204kg \cdot g[/tex]
[tex]1915N = 214kg \cdot g[/tex]
[tex]\frac{1915N}{214kg} = g[/tex]
[tex]g \approx 8.948598 m^{2}s^{-2}[/tex]
Which is not correct for the gravitational acceleration on Earth.
 
  • #6
yolo123 said:
Hi, I have an exam soon. This is urgent. Can you check my answer?

A box with mass 100kg is being pushed with enough force such that a second box with mass 2kg does not slide down and the two accelerate to the right together.
If us=0.5 and uk=0.1, what is acceleration minimum such that box m2 does not slide (please see picture.)? What is minimum value of Fpush?

What would change if [STRIKE]block[/STRIKE] box m1 is moving at constant speed?

I find that Fpush minimal is 1915N.

I find that acceleration minimal is 17.8 m/s^2.
As dauto said, please show your work.

It is possible to find the minimum acceleration needed to keep box m2 from sliding down the face of box m1. (I get a somewhat different answer than you do. Did you use g = 8.9 m/s2 rather than 9.8 m/s2 ?)

However, without some clarification, it's not possible to determine Fpush without knowing if the coefficients of friction also apply to the horizontal surface and the bottom of box m1.
If block moves at constant speed, the block would also move at constant speed in x direction but be affected in y direction because of gravity and just fall.
(The given problem calls these objects "boxes".)
That last answer looks good otherwise.


yolo123 said:
Please I know this is lots of work, but you've got to help me.
Please wait at least 24 hours before "bumping" your thread.

attachment.php?attachmentid=68662&d=1397521627.jpg
 
  • #7
Sammy S, yes there is friction between the ground, and between block one and two.

I am sorry. I have lots of exams. I got a bit carried away. I corrected some stuff. Here are my steps (my exam is in 12 hours :S). Tell me if it makes sense!
Sum of forces in x for block 1 = Fp-0.1(100)g-n=100a
Sum of forces in x for block 2=n=2a

Sum of forces in y for block 1=N-100(9.8)=0
Sum of forces in y for block 2=0.5n-2g=0

0.5n-2g=0
-0.5(n=2a)
========
2g=a
Fp-0.1(100)g-n=100*2g
Fp-0.1(100)g=204g
Fp=2098N
a= 19,6 m/s^2
 
  • #8
yolo123 said:
Sammy S, yes there is friction between the ground, and between block one and two.

I am sorry. I have lots of exams. I got a bit carried away. I corrected some stuff. Here are my steps (my exam is in 12 hours :S). Tell me if it makes sense!
Sum of forces in x for block 1 Fp-0.1(100)g-n=100a
That's not quite right.

It's really Fp-0.1(N)-n=100a .

Maybe you found N incorrectly below.
Sum of forces in x for block 2 n=2a
That is correct.
Sum of forces in y for block 1 N-100(9.8)=0
Box 1 (or Block 1) exerts an upward force of 0.5n on Box 2 (via friction). By Newton's 3rd Law, Box 2 exerts a downward force of 0.5n on Box 1. You need to include that.

In the end, that force is m2g .

That needs to be included in the vertical forces on Box 1. This will change your value for N.
Sum of forces in y for block 2=0.5n-2g=0

0.5n-2g=0
-0.5(n=2a)
========
2g=a

Fp-0.1(100)g-n=100*2g
Fp-0.1(100)g=204g
Fp=2098N
a= 19,6 m/s^2
 

1. How does the mass of the boxes affect the dynamics of box pushing another box?

The mass of the boxes plays a crucial role in determining the dynamics of box pushing another box. The heavier the boxes are, the more force is required to move them. This means that heavier boxes will have a slower acceleration and a longer stopping distance compared to lighter boxes.

2. What is the relationship between friction and the movement of the boxes?

Friction is the force that opposes the movement of objects in contact with each other. In the case of box pushing another box, friction plays a significant role in the dynamics. The amount of friction between the boxes will determine how much force is needed to move them and how fast they will accelerate.

3. Can the angle of the surface affect the dynamics of box pushing another box?

Yes, the angle of the surface can affect the dynamics of box pushing another box. When the surface is inclined, the force of gravity acting on the boxes will have a component parallel to the surface, making it harder to push the boxes. This will result in a slower acceleration and a longer stopping distance.

4. How does the force applied to the boxes affect their movement?

The force applied to the boxes will determine their acceleration and velocity. The more force applied, the faster the boxes will accelerate and the greater their velocity will be. However, if the force applied is not enough to overcome the friction between the boxes and the surface, the boxes will not move at all.

5. What is the role of inertia in box pushing another box dynamics?

Inertia is the tendency of objects to resist changes in their state of motion. In box pushing another box, the inertia of the boxes will determine how hard it is to start and stop their movement. The more massive the boxes are, the more inertia they have, and the harder it is to change their state of motion.

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