Man pulling a cube on a rough surface

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SUMMARY

The discussion focuses on calculating the maximum mass of a box that a man can pull at an angle of 10 degrees, given his mass of 70 kg and the coefficients of friction between his legs and the floor (static: 0.6) and between the box and the floor (kinetic: 0.3). The calculations reveal that the maximum mass of the box is approximately 164.87 kg. The solution involves applying the friction formula, f=μN, and considering both horizontal and vertical force components exerted by the man while pulling the box.

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Homework Statement


A man of mass 70[kg] pules a box at an angle 100. the static coefficient of friction between his legs and the floor is 0.6 and the kinetic coefficient between the box and the floor is 0.3
What's the maximum's box's mass

Homework Equations


Friction: f=μN

The Attempt at a Solution


The horizontal force the man can apply:
$$70\cdot 9.81\cdot 0.6=412.02=F\cdot \cos 10^0\rightarrow F=418.38$$
The vertical component of F raises the box: ##F\cdot \sin 10^0=418.02\cdot\sin 10^0=72.65##
the force needed to pull the box:
$$(mg-F\cdot \sin 10^0)0.3=412.02\rightarrow (mg-72.65)0.3=412.02\rightarrow m=147.4[kg]$$
It should be 164.85
 

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The vertical component of F does more than just that !
 
The vertical component also presses down on the man:
$$\left\{ \begin{array}{l} (mg-F\cdot \sin10^0)0.3=(70\cdot g+F\cdot \sin10^0)0.6 \\ F\cdot \cos10^0=(mg-F\cdot \sin10^0)0.3 \end{array} \right. $$
$$\rightarrow F=467.88,\ m=164.87\ \surd$$
 

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