Static Friction of sliding box on plane

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The discussion revolves around calculating the static friction of a small box on a larger box using the equation Fs = μ FN. The user initially miscalculated the normal force (FN) required to achieve a static friction force (Fs) of 46 N, arriving at FN = 64.8 N. After clarification, it was noted that the applied force (F) and the required acceleration of the small box were essential for the correct calculation. The user then recalculated the acceleration to be 13.8 m/s² and determined the total applied force to be 465 N. This highlights the importance of considering both the normal force and the applied force in friction problems.
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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