1. Non-Concurrent Force Systems

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3. 𝑀 𝐵 = 𝐶 𝑥 ∙9𝑐𝑚+45𝑁∙cos⁡(64°)∙36𝑐𝑚=0 +
36cm 𝑥 𝑦
45𝑁
64°
9cm
𝐵 𝑥
𝐵 𝑦
𝐶 𝑥
Class Problem: For the beam shown, find the value of the reaction forces at B and C.
64˚
45N
𝑀 𝐵 = 𝐶 𝑥 ∙9𝑐𝑚+45𝑁∙cos⁡(64°)∙36𝑐𝑚=0 +
𝐶 𝑥 = −45𝑁∙cos⁡(64°)∙36𝑐𝑚 9𝑐𝑚
𝐹 𝑥 = 𝐵 𝑥 + 𝐶 𝑥 −45𝑁∙sin⁡(64°)=0
𝐶 𝑥 =−78.91𝑁
𝐵 𝑥 −78.91𝑁 −45𝑁∙sin⁡(64°)=0
𝐵 𝑥 =78.91𝑁+45𝑁∙sin⁡(64°)
𝐹 𝑦 = 𝐵 𝑦 −45𝑁∙cos⁡(64°)=0
𝐵 𝑥 =119.36𝑁
𝐵 𝑦 =45𝑁∙cos⁡(64°)
𝐵 𝑦 =19.73𝑁

4. Class Problem: A father and son circus act are practicing their unicycle performance where they juggle and balance objects while riding a unicycle up a 28° incline. The unicycle and rider are balanced directly above point B which is 3.5m along the beam from A. The combined weight of the unicycle and both riders is 1000 N. Find the reaction at C required for equilibrium.
28°
28°
𝑊=1000𝑁
𝑊 𝑦
𝐶 𝑦
𝑊 𝑥
𝐴 𝑥
𝐴 𝑦
𝑀 𝐴 =−1000𝑁∙ cos 28° ∙3.5𝑚+ 𝐶 𝑦 ∙15𝑚=0 +
−1000𝑁∙ cos 28° ∙3.5𝑚+ 𝐶 𝑦 ∙15𝑚=0
𝐶 𝑦 = 1000𝑁∙ cos 28° ∙3.5𝑚 15𝑚 =206.02𝑁

5. 𝑊=1000𝑁
𝑊 𝑦
𝑊 𝑥
𝐴 𝑥
𝐴 𝑦
𝐶 𝑦
28°
Class Problem: Refer to the unicycle problem from the previous slide. Write Cy in terms of x, where x represents the distance travelled on the beam, and plot the equation. Develop an expression for Ay and Ax in terms of x and plot them on the same graph as Cy.
𝐶 𝑦 = 1000𝑁∙ cos 28° ∙3.5𝑚 15𝑚 ⟹ 𝐶 𝑦 (𝑥)= 1000𝑁∙ cos 28° ∙𝑥 15𝑚
𝐹 𝑦 = 𝐴 𝑦 −1000𝑁∙ cos 28° + 𝐶 𝑦 =0
𝐴 𝑦 =882.95𝑁− 𝐶 𝑦 ⟹ 𝐴 𝑦 𝑥 =882.95𝑁− 1000𝑁∙ cos 28° ∙𝑥 15𝑚
𝐹 𝑥 = 𝐴 𝑥 −1000𝑁∙sin⁡(28°)=0
𝐴 𝑥 =1000𝑁∙sin⁡(28°)=469.47N
𝐴 𝑥 (𝑥)=469.47N
The Ax value is not dependent on x, the distance travelled along the beam. It is a constant value.