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