Question

Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a force of P = 390 lb. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Answer 1

Answer:

a) Fb= 275.77 lb   Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

   Fay = 337.75 lb

   Fbx = 195 lb

   Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.

a) For no translation condition

∑ $F_{x} = 0$      &     ∑$F_{y} = 0$

Hence,

$F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0$

$F_{A}sin(30) - F_{B} sin(45) = 0$

and

$F_{A} = 390 lb$

Inserting the value of $F_{A}$ and solving the remaining equations simultaneously yields (magnitudes),

$F_{B} = 275.77 lb\\F_{C} = 142.75 lb$

b) Summing up moments

$M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))$

$M = -779.97 lb.ft$ (i.e. 779.97 lb.ft clockwise)

c)

$F_{Ax} = 390 sin(30) = 195 lb$

$F_{Ay} = 390 cos(30) = 337.75lb$

$F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb$

Answer 2

The resultant couple moment that is produced in the system of the three tug boats is; 780 lb in the clockwise direction

What are the forces applied to the couple?

1) From the attached free body diagram of the tug boat, resolving forces in the x and y direction gives;

∑fx;

F_a * cos 30 = F_b * cos 45 + F_c   ------(eq 1)

F_a * sin 30 = F_b * sin 45   -----(eq 2)

We are given F_a = 390 lb

Putting 390 lb for F_a in eq 2 gives; F_b = 275.77 lb

Also plugging in the relevant values into eq 1 gives;

F_c = 142.75 lb

2) Taking the sum of moments in both directions gives;

M = 45(-F_ay - F_cy) + 5(-F_by) + 22(-F_ax - F_by)

Plugging in the relevant vaues gives;

M ≈ -780 lb

3)  After resolving the forces at A and B into x and y components, we will have;

F_ax = 390 * sin 30 = 390 * 0.5 = 195 lb

F_ay = 390 * cos 30 = 390 * 0.8660 = 337.75 lb

F_bx = F_b * sin 45 = 275.77 * 0.7071 = 195 lb

F_by = F_b * cos 45 = 275.77 * 0.7071 = 195 lb

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