Answers
Answer:
Approximately [tex]3.5\; \rm kN[/tex] in each of the two ropes.
Explanation:
Let [tex]F_1[/tex] and [tex]F_2[/tex] 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 [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \cos(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \cos(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: [tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}}[/tex]. That force should be equal to [tex]5\; \rm kN[/tex].
The second direction is perpendicular to the resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \sin(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \sin(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: [tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}}[/tex]. 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:
[tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}} = 5\; \rm kN[/tex], and[tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}} = 0[/tex].The second equation suggests that [tex]F_1 = F_2[/tex]. Hence, replace the [tex]F_2[/tex] in the first equation with [tex]F_1[/tex] and solve for [tex]F_1\![/tex]:
[tex]F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN[/tex].
Because [tex]F_1 = F_2[/tex] (as seen in the second equation,) [tex]F_2 = F_1 \approx 3.5\; \rm kN[/tex].
In other words, the tension in each of the two ropes is approximately [tex]3.5\; \rm kN[/tex].
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 ropesThe tension in each of the ropes is calculated as follows;
The tension in vertical direction
[tex]F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N[/tex]
The tension in horizontal direction;
[tex]F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N[/tex]
Thus, the tension in each of the rope is 3,535.5 N
Learn more here:https://brainly.com/question/11194858
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