Determine The Magnitude Of The Pin Force At B

Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam.

Determine The Magnitude Of All Pin Forces B. Determine The Required Input Torque To Drive The Mechanism In This Position C. Determine The Required Input Torque Via The Energy Method Treat W=25 Lbf As A Force. Convert The Weights W2, W3, And Wa, Which Are Given In Ibf, To Slugs, E.g., M 32.2=0.1832 2-0.1832 Slugs Even Though 12, 13, And La Are Given. Determine the reactions at the supports A and B of the beam. Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.10b. Computation of reactions. The distributed loading is first replaced with a single resultant force, as seen in Figure 3.10c. The magnitude of the resultant force is. A) Call the forces acting at B Fy and Fx with to the right and up as the positive directions Sum moments about A to zero with counter clockwise moments as positive. 0 = 1000 (6 + 4) - Fy (4) + Fx.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

We can now write a moment equation at point A to determine force B_y:

circlearrowright^+ M_A=0;

4/77: Determine the magnitude of the pin force at A. Bolton 330 Engineering and Tech Engineering Mechanics: Statics 8 Edition.

600(4)+400text{cos},(15^0)(12)-B_y(12)=0
B_y=586.37 N

Now, we can write an equilibrium equation for the x-axis forces.

rightarrow^+Sigma F_x=0;

4/70 Determine The Magnitude Of The Pin Force At B

A_x-400text{sin},(15^0)=0
A_x=103.53 N

Lastly, we can write an equilibrium equation for the y-axis forces.

uparrow^+ Sigma F_y=0;
A_y+B_y-600-400text{cos},(15^0)=0

(Remember we already found B_y=586.37 N)

A_y=400 N

Now that we know A_x and A_y, we can find the magnitude. Remember, magnitude is:F_A=sqrt{(A_x)^2+(A_y)^2}
F_A=sqrt{(103.53)^2+(400)^2}