Questions: A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. The force acting at point B is (FB=49 N). For (alpha=40^circ). determine the magnitude and the location of the line of action of the equivalent force. The magnitude of the equivalent force is 9 N and the line of action P intersects line AB at m to the left of B. (Round the final answer to three decimal places.)

Solution Steps

• Given forces at points A and C are 15 N each at an angle α = 40°.
• Resolve these forces into horizontal and vertical components.

For force at A: \[ F_{Ax} = 15 \cos(40°) \] \[ F_{Ay} = 15 \sin(40°) \]

For force at C: \[ F_{Cx} = 15 \cos(40°) \] \[ F_{Cy} = 15 \sin(40°) \]

• Sum the horizontal and vertical components to find the resultant force.

Horizontal components: \[ F_{Rx} = F_{Ax} + F_{Cx} = 15 \cos(40°) + 15 \cos(40°) = 2 \times 15 \cos(40°) \]

Vertical components: \[ F_{Ry} = F_{Ay} + F_{Cy} + F_B = 15 \sin(40°) + 15 \sin(40°) + 49 = 2 \times 15 \sin(40°) + 49 \]

Resultant force: \[ F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} \]

• Calculate the moments created by the forces about point B.

Moment due to force at A: \[ M_A = 15 \sin(40°) \times 400 \text{ mm} + 15 \cos(40°) \times 240 \text{ mm} \]

Moment due to force at C: \[ M_C = 15 \sin(40°) \times 400 \text{ mm} - 15 \cos(40°) \times 240 \text{ mm} \]

Sum of moments: \[ M_{total} = M_A + M_C \]

• Use the resultant force and the total moment to find the location of the equivalent force.

\[ d = \frac{M_{total}}{F_R} \]

Final Answer