Question

An idealised frame supports a load as shown in the figure. The horizontal component of the force transferred from the horizontal member PQ to the vertical member RS at P is \underline{\hspace{2cm} N (round off to one decimal place).} \includegraphics[width=0.35\linewidth]{image99.png}

Hint:

For resolving forces in a frame, use trigonometric relationships based on the geometry of the frame to find horizontal and vertical components of the applied forces.

Answer 1

Let the force transferred from PQ to RS at P be denoted by \( F_P \). Since the system is in equilibrium, we can resolve the forces using the geometry of the frame. To find the horizontal component of the force transferred to RS, we use the method of joint resolution. The force in the horizontal member PQ (denoted as \( F_{PQ} \)) is balanced by the horizontal component in the vertical member RS, which we are solving for. Using the trigonometric relationship between the horizontal and vertical components, we have: \[ F_P = F_{PQ} \times \cos(\theta) \] Given the angle \( \theta = \tan^{-1}\left(\frac{0.3}{1.2}\right) \), the horizontal component of force is computed using the geometry and the angle from the horizontal. The horizontal component at P can be calculated as: \[ F_P = 10 \, \text{N} \times \frac{1.2}{\sqrt{(1.2^2 + 0.3^2)}} \] Now, solving the above equation: \[ F_P = 10 \times \frac{1.2}{\sqrt{1.44 + 0.09}} = 10 \times \frac{1.2}{\sqrt{1.53}} = 10 \times \frac{1.2}{1.237} \approx 18.0 \, \text{N} \] Thus, the horizontal component of the force at P is approximately \( \boxed{18.0 \, \text{N}} \).