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zysi [14]
3 years ago
9

The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

he flagpole. The other rope has a tension T, as shown. The resultant force is down the pole and of magnitude 200 N. In the space below, using a scale of 1 cm = 20 N, draw a scale drawing to find the value of the tension T. Clearly label 100 N, 200 N and T on your drawing
Physics
1 answer:
KatRina [158]3 years ago
8 0

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

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Let m denote the mass of this block. It is given that m = 1\; {\rm kg}. The weight of this block would be:

\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}.

Hence, the force that the boy applies on this block would be upward with a magnitude of F = 9.8\; {\rm N}.

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=

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