Question

A garage door is mounted on an overhead rail (given figure). The wheels at $\mathit{A}$and$\mathit{B}$have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is$\mathbf{0}\mathbf{.}\mathbf{52}$. The distance between the wheels is$\mathbf{2.00}\mathbf{\text{\hspace{0.17em}m}}$, and each is$\mathbf{\text{0.50\hspace{0.17em}m}}$from the vertical sides of the door. The door is uniform and weighs$\mathbf{\text{950N}}$. It is pushed to the left at constant speed by a horizontal force$\overrightarrow{\mathbf{F}}$

1. If the distance$\mathbf{h}$is$\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{}\mathit{m}$, what is the component of the force exerted on each wheel by the track?
2. Find the maximum value$\mathit{h}$can have without causing one wheel to leave the track.

Short answer

1. The component of the force exerted on each wheel by the track is${\mathrm{N}}_{\mathrm{A}}=80\text{\hspace{0.17em}N}$and${\mathrm{N}}_{\mathrm{B}}=870\text{\hspace{0.17em}N}$
2. The maximum value $h$can have without causing one wheel to leave the track is $1.92\text{\hspace{0.17em}m}$.

Step-by-step solution

Friction force

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

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

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

 Step 2: Identification of given data

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

Weigh of the door is$\mathrm{W}=950\text{\hspace{0.17em}N}$

The distance between the wheels is $2.00\text{\hspace{0.17em}m}$, and each is$0.50\text{\hspace{0.17em}m}$

Determine the component of the force exerted on each wheel by the track when the distance h is 1.60m.

(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,

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

role="math" localid="1668093025727" ${\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=950\text{\hspace{0.17em}N}$ (2)

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

From equation (1),

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

From equation (2),

$\begin{array}{l}\mathrm{F}={\mathrm{\mu }}_{\mathrm{k}}\left(950\text{\hspace{0.17em}N}\right)\\ \mathrm{F}=0.52\left(950\text{\hspace{0.17em}N}\right)\\ \mathrm{F}=494\text{\hspace{0.17em}N}\end{array}$

Now, to find $N_A,N_B$. 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),

$\begin{array}{c}{\mathrm{N}}_{\mathrm{B}}=950\text{\hspace{0.17em}N}-{\mathrm{N}}_{\mathrm{A}}\\ =950\text{\hspace{0.17em}N}-80\text{\hspace{0.17em}N}\\ \text{=870\hspace{0.17em}N}\end{array}$

Hence, the component of the force exerted on each wheel by the track is$N_A=80\text{\hspace{0.17em}N}$ and$N_B=870\text{\hspace{0.17em}N}$

Finding the maximum value h  can have without causing one wheel to leave the track.

(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,$N_B=W=950\text{\hspace{0.17em}N}$

Now, from equation (1)

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

Solve as:

By putting the numerical values solve as:

$\begin{array}{c}h=\frac{950\text{\hspace{0.17em}N}}{494\text{\hspace{0.17em}N}}\\ =1.92\text{\hspace{0.17em}m}\end{array}$

Hence, the maximum value $h$can have without causing one wheel to leave the track is $1.92\text{\hspace{0.17em}m}$.