Question

Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force \(\vec{P}\) of magnitude 100 N is applied at point A of the frame.

Suppose the force \(\vec{P}\) is resolved parallel to the arms AB and AC of the frame.
The magnitude of the resolved component along the arm AC is x N.
The value of x, to the nearest integer, is ___________.
[Given : \(\sin(35^\circ) = 0.573\), \(\cos(35^\circ) = 0.819\), \(\sin(110^\circ) = 0.939\), \(\cos(110^\circ) = -0.342\)]

Hint:

The arms AB and AC make 70 and 35 degrees with the vertical, so they are 35 degrees apart; resolve P along them using the sine rule on the force triangle.

Answer 1

Correct Answer: 164

From the figure, AB makes \(70^\circ\) with the wall (vertical) and AC makes \(180^\circ - 145^\circ = 35^\circ\) with the vertical. Hence P (vertical) makes \(35^\circ\) with AC and \(70^\circ\) with AB, and the angle between the arms is \(35^\circ\).
Resolve \(\vec P = \vec P_{AB} + \vec P_{AC}\) along the two arms (parallelogram law). In the force triangle, the angle opposite to \(P_{AC}\) is \(110^\circ\) and the angle opposite to \(P\) is \(35^\circ\).
Sine rule: \(\frac{P_{AC}}{\sin110^\circ} = \frac{P}{\sin35^\circ}\).
\(P_{AC} = 100\times\frac{0.939}{0.573} = 163.9\ \mathrm{N}\approx 164\ \mathrm{N}\).
Answer: 164