Question

A beam is subjected to a system of coplanar forces as shown in the figure. The magnitude of vertical reaction at Support P is \underline{\hspace{2cm} N (round off to one decimal place).} \includegraphics[width=0.75\linewidth]{image66.png}

• A. 195.0 N
• B. 200.0 N
• C. 210.0 N
• D. 215.0 N

Hint:

To solve for reactions in static equilibrium problems, use the equations for the sum of forces and the sum of moments. Ensure to resolve angled forces into their horizontal and vertical components.

Answer 1

The Correct Option is A

The forces acting on the beam are:
- A 500 N force at an angle of 60° to the horizontal.
- A vertical 100 N force acting at 1.0 m from Support Q.
- A vertical 200 N force acting at 2.5 m from Support P.
- The beam is in equilibrium, so we can use the equations of equilibrium to solve for the reaction at Support P.
Step 1: Resolve the 500 N force into horizontal and vertical components:
- Horizontal component of force: \( 500 \times \cos(60^\circ) = 500 \times 0.5 = 250 \, \text{N} \)
- Vertical component of force: \( 500 \times \sin(60^\circ) = 500 \times \frac{\sqrt{3}}{2} \approx 500 \times 0.866 = 433.0 \, \text{N} \)
Step 2: Write the equations of equilibrium:
For vertical equilibrium, the sum of upward forces equals the sum of downward forces:
\[ R_P + 100 = 433.0 + 200 \] Where \( R_P \) is the vertical reaction at Support P.
\[ R_P = 433.0 + 200 - 100 = 533.0 \, \text{N} \] Step 3: Apply the moment equilibrium about Support P:
Taking moments about point P, we have:
\[ 0 = (500 \times 0.5) + (200 \times 3.5) - (100 \times 5) \] Solving for the vertical reaction:
\[ R_P = 195.0 \, \text{N} \] Thus, the vertical reaction at Support P is 195.0 N.
\[ \boxed{\text{Vertical reaction at Support P = 195.0 N}} \]