Question

A door has a height of $\mathbf{2}\mathbf{.1}\mathbf{\text{\hspace{0.17em}m}}$ along a yaxis that extends vertically upward and a width of $\mathbf{0}\mathbf{.91}\mathbf{\text{\hspace{0.17em}m}}$along an xaxis that extends outward from the hinged edge of the door. A hinge 0.30 m from the top and a hinge 0.30 m from the bottom each support half the door’s mass, which is$\mathbf{27}\mathbf{\text{\hspace{0.17em}kg}}$ . In unit-vector notation, (a) what is the forces on the door at the top hinge and (b) what is the forces on the door at the bottom hinge?

Short answer

a) Force on the door at the top hinge in unit vector notation is,

.${\overrightarrow{F}}_{top}=(-80\widehat{i}+1.3\times {10}^2\widehat{j})\text{\hspace{0.17em}N}$

b) Force on the door at the bottom hinge in unit vector notation is,

.${\overrightarrow{F}}_{bottom}=(+80\widehat{i}+1.3\times {10}^2\widehat{j})\text{\hspace{0.17em}N}$

Step-by-step solution

Understanding the given information

i) Height of the door, $h=2.1\text{\hspace{0.17em}m}$.

ii) Width of the door, $w=0.91\text{\hspace{0.17em}m}$.

iii) Hinge distance from the top is $d_{top}=0.30\text{\hspace{0.17em}m}$.

iv) Hinge distance from the bottom is$d_{bottom}=0.30\text{\hspace{0.17em}m}$ .

v) Mass of the door, $m=27\text{\hspace{0.17em}kg}$.

Concept and formula used in the given question

You use the condition for equilibrium and write thetorque equation to find the force components of the hinge on top and at the bottom.

(a) Calculation for theforces on the door at the top hinge

You have been given that each hinge supports half the weight of the door, so the hinge’s vertical force component is,

$\begin{array}{c}{F}_y-\left(\frac{mg}{2}\right)=0\\ F_y=\frac{mg}{2}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_y=\frac{27\times 9.8}{2}\\ =132.3\text{\hspace{0.17em}N}\end{array}$

We can find the horizontal component by using the torque equation,

$\begin{array}{c}{\tau }_{net}=0\\ F_h(h-2\left(d_{bottom}\right))-mg\left(\frac{w}{2}\right)=0\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_h(2.1-2\left(0.30\right))-27\times 9.8\left(\frac{0.91}{2}\right)=0\\ F_h=80\text{\hspace{0.17em}N}\end{array}$

For equilibrium, the horizontal force component should be in the opposite direction.

So,

$F_h=-80\text{\hspace{0.17em}N}$

We can write the force in unit vector notation as,

${\overrightarrow{F}}_{top}=(-80\widehat{i}+1.3\times {10}^2\widehat{j})\text{\hspace{0.17em}N}$

Thus, force on the door at the top hinge in unit vector notation is,

.${\overrightarrow{F}}_{top}=(-80\widehat{i}+1.3\times {10}^2\widehat{j})\text{\hspace{0.17em}N}$

(b) Calculation for theforces on the door at the bottom hinge

The door would be hanging on the top hinge. But it would be resting on the bottom hinge.

Therefore, the force would remain the same, but its direction would change. So, on the bottom hinge, ${\overrightarrow{F}}_{bottom}=(+80\widehat{i}+1.3\times {10}^2\widehat{j})\text{\hspace{0.17em}N}$.