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In Fig. 1, $ angle AOC = 90^ ext{°}$ and $ angle BOC = heta^ ext{°}$ - Edexcel - A-Level Maths: Mechanics - Question 2 - 2003 - Paper 1
Question 2
In Fig. 1, $ angle AOC = 90^ ext{°}$ and $ angle BOC = heta^ ext{°}$. 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^ ext{°}$ and $ angle BOC = heta^ ext{°}$ - Edexcel - A-Level Maths: Mechanics - Question 2 - 2003 - Paper 1
Step 1
(a) the value, to one decimal place, of θ
Answer
To find the angle , we can utilize the equilibrium condition for the forces acting on point O. The force acting along OA is 8 N, and along OB is 12 N.
Using the cosine law for the triangle formed by the forces:
egin{align*}
R(&) = 8N ext{ (force along OA)}
&= 12 ext{ (force along OB)}
\angle AOB = 90^ ext{°}
\end{align*}
We find that: where eta = 41.8^ ext{°} or .
Thus, the calculated value for to one decimal place is .
Step 2