Question 1056853
.
A block of mass 4.62 kg is drawn at constant speed a distance of 1.78 m along a horizontal floor by a rope exerting a constant force 
of magnitude 1.24 N and making an angle of 11.0 degrees above the horizontal. 
Compute the coefficient of friction between the block and the floor.
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Decompose the given force &nbsp;<B>F</B>&nbsp; into the horizontal and vertical components. 
The horizontal component is equal to &nbsp;|<B>F</B>|*cos(11°), &nbsp;the vertical component is &nbsp;|<B>F</B>|*sin(11°). &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The reaction force &nbsp;<B>N</B>&nbsp; acts from the floor vertically up to the box. &nbsp;Together with 

the vertical component &nbsp;Fy&nbsp; it balances the weight: &nbsp;|<B>W</B>| = {{{F[y]}}} + |<B>N</B>|. &nbsp;So, &nbsp;|<B>N</B>| = |<B>W</B>| - {{{F[y]}}}. 


The friction coefficient is the ratio of the shear force {{{F[x]}}} to the reaction force &nbsp;|<B>N</B>|:


{{{k}}}= {{{F[x]/N}}} = {{{F[x]/(W - F[y])}}} = {{{(F*cos(11^o))/(W - F*sin(11^o))}}}.


Complete these calculations. Surely, W = mg = 4.62*9.81 Newtons.

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{{{drawing( 302, 210, 0.0, 10.1, -2.0, 5.0,
            
            line ( 0, 0, 10,    0), 
            line (10, 0,  9.7, -0.1), 
            line (10, 0,  9.7,  0.1), 
            locate(9.6,  -0.1,  x),

            line (3, 0, 6, 0), 
            line (3, 0, 3, 2),
            line (3, 2, 6, 2),
            line (6, 2, 6, 0),


        red(line  (4.5,   1,   7.0, 2.4)),
            line  (7.0,   2.4, 6.7, 2.4),
            line  (7.0,   2.4, 6.9, 2.2),
            locate(7.1,   2.7, F),

            line  (4.5,   2.5, 4.4, 2.2),
            line  (4.5,   2.5, 4.6, 2.2),
            locate(3.8,   2.6, Fy),


      green(line  (4.5,   1,   7.0, 1)),
            line  (7.0,   1,   6.7, 1.1),
            line  (7.0,   1,   6.7, 0.9),
            locate(6.7,   0.9, Fx),

      green(line  (4.5,   1, 4.5, 4)),
            line  (4.5,   4, 4.4, 3.7),
            line  (4.5,   4, 4.6, 3.7),
            locate(4.7,   3.9, N),

            locate(4.0,   3.9, y),

            line (4.5,  0.2, 2.0, 0.2),
            locate(0.2, 0.9, F_friction),
            line (2.0,  0.2, 2.3, 0.3),
            line (2.0,  0.2, 2.3, 0.1),

            line (4.5, 1, 4.5, -2),
            locate(4.8, -1.5, Weight),
            line (4.5, -2, 4.4, -1.7),
            line (4.5, -2, 4.6, -1.7),

            arc (4.5, 1, 1.6, 1.6, 330, 360),
            arc (4.5, 1, 1.8, 1.8, 330, 360),
            locate (5.4, 1.6, alpha)
)}}}

<B>Figure</B>. &nbsp;The body on the horizontal floor
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Similar problem was solved in my lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=>Using vectors to solve problems in Mechanics: Force</A>

in this site.


I simply copied and pasted the solution (the text and the Figure) from there and replaced the input data.