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Three friends tie three ropes in a knot and pull on the ropes in different directions. Adrienne (rope 1) exerts a 20-N force in the positive x-direction, and Jim (rope 2) exerts a 40-N force at an angle 53∘ above the negative x-axis. Luis (rope 3) exerts a force that balances the first two so that the knot does not move.
a. Use equilibrium conditions to find F LonKx and F LonKy.
b. Use equilibrium conditions to determine the magnitude of F⃗ LonK.
c. Use equilibrium conditions to determine the angle θ that describes the direction of F⃗ LonK. Use positive values if the force is directed above the positive x-axis and negative values if it is directed below the positive x-axis.
d. Construct a force diagram for the knot.

Respuesta :

Answer:

Part a)

[tex]F = 4 \hat i - 32\hat j[/tex]

Part b)

Magnitude of the force is given as

[tex]F = 32.2 N[/tex]

Part c)

As we know that Y component of the force is negative so here the force is directed below the X axis

[tex]\theta = -82.9 degree[/tex]

Explanation:

Part a)

Adrienne apply the force of 20 N along +X direction and Jim apply force of 40 N at 53 degree above negative X axis

so we will have

[tex]F_1 = 20 \hat i[/tex]

[tex]F_2 = 40cos53(-\hat i) + 40 sin53(\hat j)[/tex]

[tex]F_2 = -24\hat i + 32\hat j[/tex]

now let say the force exerted by Luis is F such that sum of all forces must be zero

now we have

[tex]F_1 + F_2 + F_3 = 0[/tex]

so we have

[tex]20\hat i -24\hat i + 32\hat j + F = 0[/tex]

[tex]-4\hat i + 32\hat j + F = 0[/tex]

[tex]F = 4 \hat i - 32\hat j[/tex]

Part b)

Magnitude of the force is given as

[tex]F = \sqrt{4^2 + 32^2}[/tex]

[tex]F = 32.2 N[/tex]

Part c)

As we know that Y component of the force is negative so here the force is directed below the X axis

So here we have

[tex]tan\theta = -\frac{32}{4}[/tex]

[tex]\theta = -82.9 degree[/tex]

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