Physics, 22.11.2020 14:00 laura2school

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5kN force directed along the axis of the barge, determine the tension in
each of the ropes for α = 45

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Answer from: samantha9014

The tension in each of the rope is 3,535.5 N

The given expression:

the resultant force, R = 5 kN = 5000 N

the angle in between the forces, α = 45

To find:

the tension in each of the ropes

The tension in each of the ropes is calculated as follows;

The tension in vertical direction

$F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N$

The tension in horizontal direction;

$F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N$

Thus, the tension in each of the rope is 3,535.5 N

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A barge is pulled by two tugboats. If the resultant of the forces exerted by the

tugboats is 5kN

Answer from: bellad05

Approximately $3.5\; \rm kN$ in each of the two ropes.

Explanation:

Let F_1 and F_2 denote the tension in each of the two ropes.

Refer to the diagram attached. The tension force in each rope may be decomposed in two directions that are normal to one another.

The first direction is parallel to resultant force on the barge.

The component of F_1

in that direction would be $\displaystyle F_1\cdot \cos(45^\circ) = \frac{F_1}{\sqrt{2}}$ .Similarly, the component of F_2

in that direction would be $\displaystyle F_2\cdot \cos(45^\circ) = \frac{F_2}{\sqrt{2}}$ .

These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: $\displaystyle \frac{F_1 + F_2}{\sqrt{2}}$ . That force should be equal to $5\; \rm kN$ .

The second direction is perpendicular to the resultant force on the barge.

The component of F_1

in that direction would be $\displaystyle F_1\cdot \sin(45^\circ) = \frac{F_1}{\sqrt{2}}$ .Similarly, the component of F_2

in that direction would be $\displaystyle F_2\cdot \sin(45^\circ) = \frac{F_2}{\sqrt{2}}$ .

These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: $\displaystyle \frac{F_1 - F_2}{\sqrt{2}}$ . However, the net force on the barge is normal to this direction. Therefore, the resultant force should be equal to zero.

That gives a system of two equations and two unknowns:

$\displaystyle \frac{F_1 + F_2}{\sqrt{2}} = 5\; \rm kN$ , and $\displaystyle \frac{F_1 - F_2}{\sqrt{2}} = 0$ .

The second equation suggests that F_1 = F_2 . Hence, replace the F_2 in the first equation with F_1 and solve for $F_1\!$ :

$F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN$ .

Because F_1 = F_2 (as seen in the second equation,) $F_2 = F_1 \approx 3.5\; \rm kN$ .

In other words, the tension in each of the two ropes is approximately $3.5\; \rm kN$ .

Answer from: Quest

seeing the graph would

Answer from: Quest

Aliquid is an answer i hope this

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