Static Friction of sliding box on plane

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Homework Help Overview

The discussion revolves around the static friction of a sliding box on a plane, focusing on the forces acting on the boxes involved and the calculations related to normal force and applied force.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of static friction and the normal force, with one participant expressing confusion about the relationship between the forces. Questions arise regarding the required acceleration of the boxes and the relevance of their masses in the context of the problem.

Discussion Status

Some participants have provided guidance on the need to differentiate between the normal force and the applied force. There is acknowledgment of errors in calculations, and attempts to clarify the relationships between the forces are ongoing.

Contextual Notes

There is mention of specific values for mass and coefficients of friction, but the overall setup and assumptions regarding the forces acting on the boxes are still being explored.

goonking
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Homework Statement


BaQTzvR.png


Homework Equations


Fs = μ FN

The Attempt at a Solution


Ok, so I drew a free body diagram and tilted it 90 degrees so that Fg of the small box points right and Fstatic points left.

Fg = 4.7 kg x 9.8m/s2 = 46 N

we want Fs to equal 46 N , correct? so we use the formula

Fs = (0.71) x FN

FN being the force of big box on small box

.71 x FN = 46 N

FN = 64.8N

Which is wrong. any help?
 
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goonking said:

Homework Statement


BaQTzvR.png


Homework Equations


Fs = μ FN

The Attempt at a Solution


Ok, so I drew a free body diagram and tilted it 90 degrees so that Fg of the small box points right and Fstatic points left.

Fg = 4.7 kg x 9.8m/s2 = 46 N

we want Fs to equal 46 N , correct? so we use the formula

Fs = (0.71) x FN

FN being the force of big box on small box

.71 x FN = 46 N

FN = 64.8N

Which is wrong. any help?

You calculated the normal force between the boxes, but you need the applied force, F.
 
What is the required acceleration of box one? How is the mass of box one relevant?
 
ehild said:
You calculated the normal force between the boxes, but you need the applied force, F.
oh, silly me.

so the acceleration of the small box should be 64.8 / 4.7k which is 13.8 m/s^2

we use F=ma , m being mass of both boxes together and a being 13.8 m/s^2

F = 465 N
 
goonking said:
so the acceleration of the small box should be 64.8 / 4.7k which is 13.8 m/s^2

we use F=ma , m being mass of both boxes together and a being 13.8 m/s^2

F = 465 N
Correct!
 

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