hree control rods attached to a lever ABC exert on it the fo

hree control rods attached to a lever ABC exert on it the forces P = 50 lb and Q = 21 lb as shown in the figure.

References

eBook & Resources

Section BreakChapter: 03 Rigid Bodies: Equivalent Systems of ForcesSection: 03.03 Couples and Force-Couple Systems

Problem 03.089 - Multiple forces are applied to a lever - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTSDifficulty: Medium

Problem 03.089.a - Replace the multiple forces with an equivalent force-couple system

Replace the three forces with an equivalent force-couple system at B. (Round the moment to two decimal places and the force and its angle to the nearest whole number.) (You must provide an answer before moving to the next part.)

M =  lb•in. (Clockwise)

Solution

>> As, Both Q Forces are opposite to each other and also equal in magnitude. So, they will not produce any net force, they will only provide a moment of (Qcos20 * 70 in)

>> Here, i use Q*cos 20, because on resolving it along the lever and perpandicular to lever, component along the lever, i.e. Q*sin20 is along the line and thus, will not produce any moment.

>> So, M1 = Q*cos20*70 = 1381.348 lb-in (clockwise)

>> Now, Considering Force P.

As it is acting at an angle 55 degree perpandicular to lever and as lever, itself is at an angle 30 from horizontal.

So, Using Geometry, P is at angle 35 + 30 = 65 degree from horizontal

>> Now, Component of P perpandicular to lever = P*cos55 = 50*cos55 = 28.679 lb

So, due to it, Momen at B, M2 = 30*28.679 = 860.365 lb-in (antilockwise)

As, COmponent of P along lever will not produce any moment

SO, Net Moment at B due to all foces in the system = M1 + M2 = 1381.348 - 860.365 = 520.983 lb-in

>> SO, Net Force = P = 50 lb, at an angle 65 degree with horizontal

and, Net Momnet = 520.983 lb-in clockwise............ANSWER........