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Friction

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\n

Knowing that $\\theta=\\var{th}^\\circ$, determine the range of values of $P$ for which equilibrium is maintained.

", "advice": "

$+\\!\\!\\nearrow\\sum F_y=0,\\quad\\implies\\quad N-800 \\cos 25^\\circ+P\\sin(\\theta-25^\\circ)=0$

\n

$N=800 \\cos 25^\\circ-P\\sin( \\var{th-25}^\\circ),$

\n

$F_m=\\mu_s N$

\n

To start moving the block up the incline:

\n

\n

\n

$+\\!\\!\\searrow\\sum F_x=0,\\quad\\implies\\quad \\mu_s N-P\\cos(\\theta-25^\\circ)+800\\sin 25^\\circ=0$

\n

$\\mu_s (800 \\cos 25^\\circ-P\\sin\\var{th-25}^\\circ)-P\\cos\\var{th-25}^\\circ+800\\sin 25^\\circ=0$

\n

$\\mu_s 800 \\cos 25^\\circ-\\mu_s P\\sin\\var{th-25}^\\circ-P\\cos\\var{th-25}^\\circ+800\\sin 25^\\circ=0$

\n

$\\mu_s 800 \\cos 25^\\circ+800\\sin 25^\\circ=+P[\\mu_s \\sin\\var{th-25}^\\circ+\\cos\\var{th-25}^\\circ]$

\n

$P=\\dfrac{800(\\sin 25^\\circ+\\mu_s\\cos 25^\\circ)}{\\cos\\var{th-25}^\\circ+\\mu_s \\sin\\var{th-25}^\\circ}=\\var{Pmax}\\,\\text{N}\\quad\\blacktriangleleft$

\n

\n

To prevent the block from moving down:

\n

\n

$+\\!\\!\\searrow\\sum F_x=0,\\quad\\implies\\quad -\\mu_s N-P\\cos(\\theta-25^\\circ)+800\\sin 25^\\circ=0$

\n

$-\\mu_s (800 \\cos 25^\\circ-P\\sin \\var{th-25}^\\circ)-P\\cos\\var{th-25}^\\circ+800\\sin 25^\\circ=0$

\n

$-\\mu_s 800  \\cos 25^\\circ +\\mu_s P\\sin \\var{th-25}^\\circ- P\\cos \\var{th-25}^\\circ+800\\sin 25^\\circ=0$

\n

$\\mu_s P\\sin \\var{th-25}^\\circ- P\\cos\\var{th-25}^\\circ-\\mu_s 800  \\cos 25^\\circ +800\\sin 25^\\circ=0$

\n

$-\\mu_s 800  \\cos 25^\\circ +800\\sin 25^\\circ=P[\\cos\\var{th-25}^\\circ-\\mu_s \\sin\\var{th-25}^\\circ]$

\n

$P=\\dfrac{800(\\sin 25^\\circ-\\mu_s\\cos 25^\\circ)}{\\cos\\var{th-25}^\\circ-\\mu_s \\sin\\var{th-25}^\\circ}=\\var{Pmin}\\,\\text{N}\\quad\\blacktriangleleft$

\n

\\[\\var{Pmin}\\,\\text{N} \\le P \\le \\var{Pmax}\\,\\text{N}\\quad\\blacktriangleleft\\]

\n
\n

Alternative solution:

\n

Combine the normal and friction forces $\\vec{R}=\\vec{N}+\\vec{F}_\\mu$. When $F_\\mu=\\mu_sN$, the angle between the resultant $\\vec{R}$ and the normal force is the friction angle:

\n

\\[\\phi_s=\\arctan \\mu_s=\\arctan \\var{mu_s}=\\var{phi}^\\circ.\\]

\n

To prevent the block from moving up:

\n

\n

In this case there are only three forces acting on the block, which in view of equilibrium, form a triangle:

\n

\n

In this triangle, the angle between the force $\\vec{R}$ and weight $\\vec{W}$ is $\\phi_s+25^\\circ$. Using the law of sines:

\n

$\\displaystyle\\frac{P}{\\sin (\\phi_s+25^\\circ)}=\\frac{W}{\\sin \\left[180^\\circ-(\\phi_s+25^\\circ)-(90^\\circ-\\theta)\\right]},$

\n

$\\displaystyle\\frac{P}{\\sin\\var{phi+25}^\\circ}=\\frac{800}{\\sin\\var{180-phi-25-90+th}^\\circ},$

\n

$P=800\\dfrac{\\sin\\var{phi+25}^\\circ}{\\sin\\var{180-phi-25-90+th}^\\circ}=\\var{Pmax}\\,\\text{N}.\\,\\blacktriangleleft$

\n

\n

To prevent the block from moving down send the friction force up the incline:

\n

\n

Form a triangle:

\n

\n

In this triangle, the angle between the force $\\vec{R}$ and weight $\\vec{W}$ is $25^\\circ-\\phi_s$. Using the law of sines:

\n

$\\displaystyle\\frac{P}{\\sin (25^\\circ-\\phi_s)}=\\frac{W}{\\sin \\left[180^\\circ-(25^\\circ-\\phi_s)-(90^\\circ-\\theta)\\right]},$

\n

$\\displaystyle\\frac{P}{\\sin\\var{-phi+25}^\\circ}=\\frac{800}{\\sin\\var{180+phi-25-90+th}^\\circ},$

\n

$P=800\\dfrac{\\sin\\var{-phi+25}^\\circ}{\\sin\\var{180+phi-25-90+th}^\\circ}=\\var{Pmin}\\,\\text{N}\\,\\blacktriangleleft$

\n

\\[\\var{Pmin}\\,\\text{N} \\le P \\le \\var{Pmax}\\,\\text{N}.\\quad\\blacktriangleleft\\]

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