Answered

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

Answer :

jacob193

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].

${teks-lihat-gambar} jacob193
onyebuchinnaji

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

[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

Other Questions