Name: Equilibrium
Uploaded: 2019-07-19
Duration: 9 min 40 s
Description: The fundamental difference between a machine and a structure is the machines are dynamic while the structures are static. A study of equilibrium is essential...

Question

In the figure shown below, a uniform plank, with a length $L$ of $10.5 m$ and a weight of $445 N$ , rests on the ground and against a frictionless roller, at the top of a wall of height $h = 3.05 m$ . The plank remains in equilibrium for any value of $θ \geq 70 °$ but slips if $θ < 70 °$ . Find the coefficient of static friction between the plank and the ground. (Take $\sin 70 ° = 0.94$ and $\cos 70 ° = 0.34$ )

Accepted Answer

Let us draw a free-body diagram of the plank, as shown below.

Because the roller is frictionless, the force it exerts is normal to the plank (say $F$ ) and makes an angle $θ$ with the vertical, weight of the plank, normal, friction at the contact point of the plank and the horizontal surface.

$W$ is the force of gravity, which acts at the center of the plank, at a distance $\frac{L}{2}$ from the point where the plank touches the floor. $F_{N}$ is the normal force of the floor and $f$ is the force of friction. Let the distance from the foot of the plank to the wall be $θ \geq 70 °$ . This quantity is not given directly but it can be computed using $d = \frac{h}{tanθ}$ .

The equations of equilibrium are:

Horizontal force components: $F sinθ - f = 0$

Vertical force components: $F cosθ - W + F_{N} = 0$

Torques: $F_{N} d - f h - W (d - \frac{L}{2} cosθ) = 0$

For the torque equation, the point of contact between the plank and the roller was used as the origin.

When $θ = 70 °$ , the plank just begins to slip and $f = μ_{s} F_{N}$ , where $μ_{s}$ is the coefficient of static friction. We will use the equations of equilibrium to compute $F_{N}$ and $f$ for $θ = 70 °$ , then use $μ_{s} = \frac{f}{F_{N}}$ to compute the coefficient of friction.

The second equation gives $F = \frac{(W - F_{N})}{cosθ}$ . Substituting this in the first equation, we get,

$f = (W - F_{N}) \frac{sinθ}{cosθ} = (W - F_{N}) tanθ$

Now, substituting this in the third equation and solving the result for $F_{N}$ ,

$F_{N} = \frac{d - (\frac{L}{2}) cosθ + h tanθ}{d + h tanθ} W = \frac{h (1 + \tan^{2} θ) - (\frac{L}{2}) sinθ}{h (1 + \tan^{2} θ)} W$ ,

Here, we have used $d = \frac{h}{tanθ}$ and multiplied both numerator and denominator by $tanθ$ .

On using the trigonometric identity $1 + \tan^{2} θ = \frac{1}{\cos^{2} θ}$ and multiplying both numerator and denominator by $\cos^{2} θ$ , we get,

$F_{N} = W (1 - \frac{L}{2 h} \cos^{2} θsinθ)$

Now, we use this expression of $F_{N}$ in $f = (W - F_{N}) tanθ$ to find the friction,

$f = \frac{W L}{2 h} \sin^{2} θcosθ$ .

Substituting these expressions of $f$ and $F_{N}$ in $μ_{s} = \frac{f}{F_{N}}$ gives

$μ_{s} = \frac{L \sin^{2} θcosθ}{2 h - L {sinθcos}^{2} θ}$

Evaluating this expression for $θ = 70 °, L = 10.5 m$ and $h = 3.05 m$ gives,

$μ_{s} = \frac{(10.5 m) \sin^{2} 70 ° \cos 70 °}{2 (3.05 m) - (10.5 m) \sin 70 ° \cos^{2} 70 °} = 0.305$ .

Particle	$g$	Particle	$g$
$1$	$8.00$	$4$	$7.40$
$2$	$7.80$	$5$	$7.60$
$3$	$7.60$	$6$	$μ_{s} = \frac{(10.5 m) \sin^{2} 70 ° \cos 70 °}{2 (3.05 m) - (10.5 m) \sin 70 ° \cos^{2} 70 °} = 0.305$

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