Kainui
  • Kainui
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
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kainui
  • 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?
Kainui
  • 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.
ParthKohli
  • ParthKohli
Alright, just draw free-body diagrams and wish for the best.|dw:1433696058786:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Kainui
  • 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.
ParthKohli
  • 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).
Michele_Laino
  • 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
ParthKohli
  • ParthKohli
\[N_1 = 200 g\tag{top box, vertical direction}\]Where \(g\) is the gravitational constant.\[N_1 + 200 g = N_2 \tag{bottom box, vertical direction}\]
ParthKohli
  • ParthKohli
Yeah, looks like my work's correct so far...
Michele_Laino
  • Michele_Laino
\[\Large {R_2} = {\mu _{SF}}2mg\]
ParthKohli
  • ParthKohli
\[N_1 = 200 g = (200 \times 9.81) ~N\]\[N_2 = 400 g = (400 \times 9.81)~N\]\[f_1 = \mu_sN_1\]\[f_2 = \mu_s' N_2\]
ParthKohli
  • ParthKohli
Again, careful about the units. I've just replaced lbs with kg because, well, SI-master-race.
ParthKohli
  • ParthKohli
For the box to move, \(F = f_1 + \epsilon\). Minimise this force and you get \(F = f_1 = \mu_s N_1\).
Michele_Laino
  • Michele_Laino
now the equilibrium condition is given by these subsequent formulas: \[\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{O}} \times {\mathbf{XP}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
|dw:1433696867479:dw|
Michele_Laino
  • Michele_Laino
oops, there is a typo: \[\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{OX}} \times {\mathbf{P}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
the first vector equation, is equivalent to this subsequent scalar equation: \[\Large P - {R_2} = 0\]
Michele_Laino
  • Michele_Laino
where I have used the reference system in my drawing
Kainui
  • 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.
Michele_Laino
  • Michele_Laino
whereas the second vector equation, is equivalent to this scalar equation: \[\Large 5P - 4.5{R_1} - 4.5{R_1} = 0\]
Michele_Laino
  • Michele_Laino
From the firs scalar equation, we see that, motion is possible, if and only if, the subsequent condition holds: \[\Large P > {R_2} = {\mu _{SF}}2mg\]
Michele_Laino
  • Michele_Laino
first*
ParthKohli
  • ParthKohli
Oh, I thought the question was talking about the motion of the first block only.
Michele_Laino
  • Michele_Laino
in that case the condition is: \[\Large {\mu _{SB}}mg < P < {\mu _{SF}}2mg\]
ParthKohli
  • ParthKohli
|dw:1433698078430:dw|