91Ó°ÊÓ

Friction force is a force that prevents two solid objects from rolling or sliding over one another and is given by,

$\mathbf{F}\mathbf{=}{\mathbf{μ}}_{\mathbf{K}}\mathbf{N}$ (1)

Where $\mathbf{F}$is friction force,${\mathbf{μ}}_{\mathbf{K}}$ is coefficient of friction $\mathbf{N}$is normal force.

Here we have, the coefficient of kinetic friction is${\mathrm{μ}}_{\mathrm{k}}=0.52$

Weigh of the door is$\mathrm{W}=950\text{ N}$

The distance between the wheels is $2.00\text{″¾}$, and each is$0.50\text{″¾}$

(a)

We have,$N_A$is normal force at point$A$and$N_B$is normal force at $B$.

Also, friction at each point of them by$f_{kA}$and $f_{kB}$.

Now, by second condition of equilibrium, the summation of force in y-direction is zero, so we get,

$\begin{array}{c}∑{\mathrm{F}}_{\mathrm{y}}=0\\ {\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}−\mathrm{W}=0\\ {\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=\mathrm{W}\end{array}$

role="math" localid="1668093025727" ${\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=950\text{ N}$ (2)

Now, for the summation of force in x-direction solve as:

From equation (1),

$\mathrm{F}={\mathrm{μ}}_{\mathrm{k}}({\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}})$

From equation (2),

$\begin{array}{l}\mathrm{F}={\mathrm{μ}}_{\mathrm{k}}\left(950\text{ N}\right)\\ \mathrm{F}=0.52\left(950\text{ N}\right)\\ \mathrm{F}=494\text{ N}\end{array}$

Now, to find $N_A,\text{ }{N}_B$. From first condition of equilibrium, summation of torque around point $B$is zero. So, we get,

Now, by substituting the values in above equation solve as:

Now, from equation (2),

$\begin{array}{c}{\mathrm{N}}_{\mathrm{B}}=950\text{ N}−{\mathrm{N}}_{\mathrm{A}}\\ =950\text{ N}−80\text{ N}\\ \text{=870 N}\end{array}$

Hence, the component of the force exerted on each wheel by the track is$N_A=80\text{ N}$ and$N_B=870\text{ N}$

(b)

At maximum $h$, the normal force at point$A$is zero and friction force at that point is also zero.

So, the normal force at point $B$is equal to weight.

So,$N_B=W=950\text{ N}$

Now, from equation (1)

Now, the summation of torque around point $B$is zero.

Solve as:

By putting the numerical values solve as:

$\begin{array}{c}h=\frac{950\text{ N}}{494\text{ N}}\\ =1.92\text{″¾}\end{array}$

Hence, the maximum value $h$can have without causing one wheel to leave the track is $1.92\text{″¾}$.