# Box pushing another box Dynamics

1. Apr 14, 2014

### yolo123

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. Apr 14, 2014

### yolo123

Please I know this is lots of work, but you've gotta help me.

3. Apr 14, 2014

### dauto

4. Apr 14, 2014

### mafagafo

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. Apr 15, 2014

### mafagafo

I think that 1915N is wrong.
$$(1915N-0.1 \cdot 100kg \cdot g ) \cdot \frac{2}{102} \cdot 0.5 = 2kg \cdot g$$
$$1915N-10kg \cdot g = 204kg \cdot g$$
$$1915N = 214kg \cdot g$$
$$\frac{1915N}{214kg} = g$$
$$g \approx 8.948598 m^{2}s^{-2}$$
Which is not correct for the gravitational acceleration on Earth.

6. Apr 15, 2014

### SammyS

Staff Emeritus
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.
(The given problem calls these objects "boxes".)
That last answer looks good otherwise.

Please wait at least 24 hours before "bumping" your thread.

7. Apr 15, 2014

### yolo123

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. Apr 15, 2014

### SammyS

Staff Emeritus
That's not quite right.

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

Maybe you found N incorrectly below.
That is correct.
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.