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Three forces F₁, F₂, and F₃ acting on a particle P are given by F₁ = (7i - 9j) N F₂ = (5i + 6j) N F₃ = (pi + qj) N where p and q are constants - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1
Question 3
Three forces F₁, F₂, and F₃ acting on a particle P are given by F₁ = (7i - 9j) N F₂ = (5i + 6j) N F₃ = (pi + qj) N where p and q are constants. Given that P... show full transcript
Worked Solution & Example Answer:Three forces F₁, F₂, and F₃ acting on a particle P are given by F₁ = (7i - 9j) N F₂ = (5i + 6j) N F₃ = (pi + qj) N where p and q are constants - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1
Step 1
find the value of p and the value of q.
Answer
For Particle P to be in equilibrium, the net force acting on it must be zero. Therefore, we sum up the components of the forces:
In the x-direction:
[ 7 + 5 + p = 0 ]
This simplifies to:
[ p = -12 ]
In the y-direction:
[ -9 + 6 + q = 0 ]
This simplifies to:
[ q = 3 ]
Thus, the values are:
[ p = -12, \quad q = 3 ]
Step 2
the magnitude of R
Answer
To find the resultant R of forces F₂ and F₃ after removing F₁, we first express F₂ and F₃ with their respective values for p and q:
[ F₂ = (5i + 6j) ] [ F₃ = (-12i + 3j) ]
Now, we can sum these:
[ R = F₂ + F₃ = (5i + 6j) + (-12i + 3j) = (-7i + 9j) ]
Next, we calculate the magnitude of R:
[ |R| = \sqrt{(-7)^2 + (9)^2} = \sqrt{49 + 81} = \sqrt{130} \approx 11.4 \text{ N} ]
Step 3
the angle, to the nearest degree, that the direction of R makes with j.
Answer
To determine the angle θ that the vector R makes with the positive y-axis (j-direction), we use the tangent function:
[ \tan \theta = \frac{|\text{component of } i|}{|\text{component of } j|} ]
Substituting our components:
[ \tan \theta = \frac{|-7|}{|9|} = \frac{7}{9} ]
Now, calculating θ:
[ \theta = \tan^{-1}\left( \frac{7}{9} \right) \approx 40.6^{\circ} ]
Thus, the angle θ to the nearest degree is:
[ \theta \approx 41^{\circ} ]