## Kainui one year ago If each box weighs 200lb , determine the least horizontal force P that the man must exert on the top box in order to cause motion. The coefficient of static friction between the boxes is μs = 0.80, and the coefficient of static friction between the box and the floor is μ′s = 0.55. https://session.masteringengineering.com/problemAsset/1526808/1/Probs.8-57_58.jpg

1. Kainui

So is the whole thing going to slide, is the top box going to fall off, or is the whole thing going fall over?

2. Kainui

This is a question my little brother sent me last semester who's in school to become an engineer. He said it was a fun problem, and I sorta forgot about it but just found it and now it's bugging me since I don't really know what to do.

3. ParthKohli

Alright, just draw free-body diagrams and wish for the best.|dw:1433696058786:dw|

4. Kainui

I guess my confusion lies in the fact that I can sorta consider the whole thing as having a kind of pivot point at point B in the diagram, but I can also consider it to be a completely solid object that slides, or I can consider the same scenario for the box on top as either pivoting or sliding. I guess I could work out all 4 or at least get started but that just doesn't seem like the right way to think about it, mindlessly calculating.

5. ParthKohli

Whenever considering limiting cases, $$f_s = \mu_s N$$ (actually, there is an inequality to the left but we consider the equality to anlayse limiting cases).

6. Michele_Laino

we have these forces: |dw:1433696128210:dw| on box A are acting these two forces: the force P and the friction force R_1 whose magnitudfe is: $\Large {R_1} = {\mu _{SB}}mg$ where $\Large {\mu _{SB}}$ is the static friction coefficient between boxes. On box B are acting the se two forces, -R_1 and R_2 whose magnitude is: \\Large [{R_2} = {\mu _{SF}}2mg\] where: $\Large {\mu _{SF}}$ is the static friction coefficient between box and floor

7. ParthKohli

$N_1 = 200 g\tag{top box, vertical direction}$Where $$g$$ is the gravitational constant.$N_1 + 200 g = N_2 \tag{bottom box, vertical direction}$

8. ParthKohli

Yeah, looks like my work's correct so far...

9. Michele_Laino

$\Large {R_2} = {\mu _{SF}}2mg$

10. ParthKohli

$N_1 = 200 g = (200 \times 9.81) ~N$$N_2 = 400 g = (400 \times 9.81)~N$$f_1 = \mu_sN_1$$f_2 = \mu_s' N_2$

11. ParthKohli

Again, careful about the units. I've just replaced lbs with kg because, well, SI-master-race.

12. ParthKohli

For the box to move, $$F = f_1 + \epsilon$$. Minimise this force and you get $$F = f_1 = \mu_s N_1$$.

13. Michele_Laino

now the equilibrium condition is given by these subsequent formulas: $\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{O}} \times {\mathbf{XP}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.$

14. Michele_Laino

|dw:1433696867479:dw|

15. Michele_Laino

oops, there is a typo: $\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{OX}} \times {\mathbf{P}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.$

16. Michele_Laino

the first vector equation, is equivalent to this subsequent scalar equation: $\Large P - {R_2} = 0$

17. Michele_Laino

where I have used the reference system in my drawing

18. Kainui

Sorry for the delay, give me a moment to read through this I'm at the kitchen table on my laptop and I'm having a conversation with my parents so it'll be a minute.

19. Michele_Laino

whereas the second vector equation, is equivalent to this scalar equation: $\Large 5P - 4.5{R_1} - 4.5{R_1} = 0$

20. Michele_Laino

From the firs scalar equation, we see that, motion is possible, if and only if, the subsequent condition holds: $\Large P > {R_2} = {\mu _{SF}}2mg$

21. Michele_Laino

first*

22. ParthKohli

Oh, I thought the question was talking about the motion of the first block only.

23. Michele_Laino

in that case the condition is: $\Large {\mu _{SB}}mg < P < {\mu _{SF}}2mg$

24. ParthKohli

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