Three control rods attached to a lever A B C exert on it the forces shown. (a) R

Question

Three control rods attached to a lever A B C exert on it the forces shown.

(a) Replace the three forces with an equivalent force-couple system at B.

(b) Determin the single force that is equivalent to the force couple system obtained in part (a) and specify it point of application on the lever.

Do not round intermediate calculations, however for display purposes report intermediate steps rounded to four significant figures. Give your final answers to three significant figures. 57 554 lb 55.4 lb 20 in. 30° 22.0 lb 20°46 in. 20° 22.0 lb Three control rods attached to a lever ABC exert on it the forces shown. Part 1 out of 2 (a ) Replace the three forces with an equivalent force-couple system at B F=

Explanation / Answer

a)

F1 = 55.4*(cos(30 + 33) i + sin(30 + 33) j) <------ in vector form

F2 = 22*(cos(90 - 20 - 30) i - sin(90 - 20 - 30) j)

Similalrly, F3 = 22*(-cos(90 - 30 - 20) i + sin(90 - 30 - 20) j )

So, net force, Fnet = F1 + F2 + F3

We can clearly see that F2 = -F3

So, Fnet = F1 + F2 - F2 = F1 = 55.4*(cos(63 deg) i + sin(63 deg) j)

= ( 25.2 i + 49.4 j ) lb

So, equivalent force magnitude = 55.4 lb <------- answer

direction = 63 deg counterclockwise from +x axis <-------answer

Couple of F2 and F3 = 22*66 = 1254 lb.in (clockwise direction)

Couple of F1 = 55.4*cos(57 deg)*20 = 603.5 lb.in (counterclockwise direction)

So, net couple = 1254 - 603.5 = 650.5 lb.in (in clockwise direction) <------- answer

b)

So, the equivalent force = 55.4 lb at 63 deg in counterclockwise direction

and it acts at position = 650.5/(55.4*cos(57 deg)) = 21.56 in from the point B along BC

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