A uniform door of width $$\(w\)$$ is supported by two hinges, X and Y . X and Y are a vertical distance $$\(L\)$$ apart as shown. The diagram shows the force acting on the door at each hinge. Not to scale The mass of the door is 14.4 kg . The weight of the door causes a moment about each hinge. (i) Show that the horizontal component of the force of hinge Y on the door is about 40 N . $$\(\begin{aligned} L & =1.60 \mathrm{~m} \\ w & =0.85 \mathrm{~m}\end{aligned}\)$$ (5) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. (ii) The vertical component of the force of each hinge on the door is equal to half the weight of the door. Hinge Y exerts a force $$\(F\)$$ on the door at an angle $$\(\theta\)$$ to the vertical, as shown. Not to scale Determine the force $$\(F\)$$ and the angle $$\(\theta\)$$. (4) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. $$\[ F= \]$$ ................................................... $$\(\theta=\)$$ ................................................... (iii) An identical door is supported by hinges that are much closer together. Explain how a smaller value of $$\(L\)$$ affects $$\(F\)$$ and $$\(\theta\)$$. You should consider moments about point X . (4) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. ..............................................................................................................................................................................................................................................
Exam No：wph11-01-que-20240112 Year：2024 Question No：18(c)
Answer:
Use of \(W=m g\)
Horizontal distance from edge of door to centre of gravity identified as \(\frac{w}{2}\)
Use of moment of force \(=F x\)
Use of clockwise moment = anticlockwise moment
Horizontal component of force of hinge Y on door \(=37.5(\mathrm{~N})\)
OR
Use of \(W=m g\)
Use of \(\tan \theta\) with dimensions of door
Use of vertical component of force of hinge Y on door \(=\mathrm{W} / 2\)
Use of \(\tan \theta\) with vertical and horizontal components of force of hinge Y on door
Horizontal component of force of hinge Y on door \(=37.5(\mathrm{~N})\)
Example of calculation
\(W=14.4 \mathrm{~kg} \times 9.81 \mathrm{~N} \mathrm{~kg}^{-1}=141 \mathrm{~N}\)
\(141 \mathrm{~N} \times \frac{0.85 \mathrm{~m}}{2}=F_{\text {horizontal }} \times 1.60 \mathrm{~m}\)
\(F_{\text {horizontal }}=\frac{60.0 \mathrm{~N} \mathrm{~m}}{1.60 \mathrm{~m}}=37.5 \mathrm{~N}\)
Use of appropriate trigonometry to determine \(\theta\)
\(\theta=28^{\circ}\) (allow ecf from 18(c)(i)) (show that answer gives \(30^{\circ}\) )
Use of Pythagoras' theorem to determine \(F\)
Or
Use of appropriate trigonometry to determine \(F\)
\(\mathrm{F}=80 \mathrm{~N}\) (allow ecf from 18(c)(i)) (show that answer gives 81 N )
Example of calculation
\[
\theta=\tan ^{-1}\left(\frac{37.5}{\left(\frac{141}{2}\right)}\right)=28.0^{\circ}
\]
\[
F=\sqrt{(37.5 \mathrm{~N})^{2}+\left(\frac{141 \mathrm{~N}}{2}\right)^{2}}=80.0 \mathrm{~N}
\]
Moment of weight of door (about X ) remains the same. Or \(m g \times \frac{w}{2}=F_{\text {horizontal }} \times L(\) and \(m, g\) and \(w\) remain the same \()\).
(So) horizontal (component of) force increases (when \(L\) decreases).
Vertical (component of) force remains the same.
Or
Each hinge still holds half the weight of the door.
(So) \(F\) increases and \(\theta\) increases.
Horizontal distance from edge of door to centre of gravity identified as \(\frac{w}{2}\)
Use of moment of force \(=F x\)
Use of clockwise moment = anticlockwise moment
Horizontal component of force of hinge Y on door \(=37.5(\mathrm{~N})\)
OR
Use of \(W=m g\)
Use of \(\tan \theta\) with dimensions of door
Use of vertical component of force of hinge Y on door \(=\mathrm{W} / 2\)
Use of \(\tan \theta\) with vertical and horizontal components of force of hinge Y on door
Horizontal component of force of hinge Y on door \(=37.5(\mathrm{~N})\)
Example of calculation
\(W=14.4 \mathrm{~kg} \times 9.81 \mathrm{~N} \mathrm{~kg}^{-1}=141 \mathrm{~N}\)
\(141 \mathrm{~N} \times \frac{0.85 \mathrm{~m}}{2}=F_{\text {horizontal }} \times 1.60 \mathrm{~m}\)
\(F_{\text {horizontal }}=\frac{60.0 \mathrm{~N} \mathrm{~m}}{1.60 \mathrm{~m}}=37.5 \mathrm{~N}\)
Use of appropriate trigonometry to determine \(\theta\)
\(\theta=28^{\circ}\) (allow ecf from 18(c)(i)) (show that answer gives \(30^{\circ}\) )
Use of Pythagoras' theorem to determine \(F\)
Or
Use of appropriate trigonometry to determine \(F\)
\(\mathrm{F}=80 \mathrm{~N}\) (allow ecf from 18(c)(i)) (show that answer gives 81 N )
Example of calculation
\[
\theta=\tan ^{-1}\left(\frac{37.5}{\left(\frac{141}{2}\right)}\right)=28.0^{\circ}
\]
\[
F=\sqrt{(37.5 \mathrm{~N})^{2}+\left(\frac{141 \mathrm{~N}}{2}\right)^{2}}=80.0 \mathrm{~N}
\]
Moment of weight of door (about X ) remains the same. Or \(m g \times \frac{w}{2}=F_{\text {horizontal }} \times L(\) and \(m, g\) and \(w\) remain the same \()\).
(So) horizontal (component of) force increases (when \(L\) decreases).
Vertical (component of) force remains the same.
Or
Each hinge still holds half the weight of the door.
(So) \(F\) increases and \(\theta\) increases.
Knowledge points:
CH1 - Mechanics
Solution:
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