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In Fig. 1, \(\angle AOC = 90^\circ\) and \(\angle BOC = \theta^\circ\) - Edexcel - A-Level Maths Mechanics - Question 2 - 2003 - Paper 1
Question 2
In Fig. 1, \(\angle AOC = 90^\circ\) and \(\angle BOC = \theta^\circ\). A particle at O is in equilibrium under the action of three coplanar forces. The three forces... show full transcript
Worked Solution & Example Answer:In Fig. 1, \(\angle AOC = 90^\circ\) and \(\angle BOC = \theta^\circ\) - Edexcel - A-Level Maths Mechanics - Question 2 - 2003 - Paper 1
Step 1
Calculate the value, to one decimal place, of \(\theta\)
Answer
To find (\theta), we can apply the equilibrium condition for forces. Since the forces are coplanar and in equilibrium, we can use the following relationship:
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Set the resultant force in the vertical direction to zero:
[ R = 12 \cos(\beta) + 8 ]
Here, (\beta = 90^\circ - \theta). So, we can rewrite it:
[ 8 = 12 \cos(\beta) + 8 ]
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Rearranging gives:
[ 0 = 12 \cos(\beta) ]
Which does not apply directly, so we get:
[ R = 12 \cos(\beta) = 8 ]
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Using the trigonometric identity, we have:
[ 12 \cos(\beta) = 8 ]
Solving for (\beta):
[ \cos(\beta) = \frac{8}{12} = \frac{2}{3} ]
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Calculate (\beta):
[ \beta = \cos^{-1}(\frac{2}{3}) \approx 41.8^\circ ]
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Then, since (\theta = 90^\circ - \beta):
[ \theta \approx 90^\circ - 41.8^\circ = 48.2^\circ ]
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Therefore, to one decimal place, (\theta \approx 48.2^\circ).
Step 2
Calculate the value, to 2 decimal places, of \(X\)
Answer
To find (X), we continue using the equilibrium of forces. We can express (X) in terms of the known forces:
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The equation for force in the horizontal direction is:
[ R = X = 12 \cos(\beta) ]
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Substituting in (\beta \approx 41.8^\circ):
[ X = 12 \cos(41.8^\circ)]
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We can calculate this:
Using (\cos(41.8^\circ) \approx 0.75"),
[ X = 12 \cdot 0.75 = 9 ]
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Thus, to two decimal places, (X \approx 8.94 N).