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Math  /  Algebra

Question20. A telephone pole has three cables pulling as shown from above, with F⃗1=(300.0i^+500.0j^),F⃗2=−200.0i^\vec{F}_{1}=(300.0 \hat{i}+500.0 \hat{j}), \vec{F}_{2}=-200.0 \hat{i}, and F⃗3=−800.0j^\vec{F}_{3}=-800.0 \hat{j}. (a) Find the net force on the telephone pole in component form. (b) Find the magnitude and direction of this net force.

Studdy Solution

STEP 1

1. The forces are given in vector form with components along the i^ \hat{i} and j^ \hat{j} directions.
2. We are required to find the net force in component form.
3. We need to calculate the magnitude and direction of the net force.

STEP 2

1. Calculate the net force in component form.
2. Calculate the magnitude of the net force.
3. Calculate the direction of the net force.

STEP 3

Calculate the net force in component form by summing up the individual force vectors.
Given: F⃗1=(300.0i^+500.0j^) \vec{F}_{1} = (300.0 \hat{i} + 500.0 \hat{j}) F⃗2=(−200.0i^) \vec{F}_{2} = (-200.0 \hat{i}) F⃗3=(−800.0j^) \vec{F}_{3} = (-800.0 \hat{j})
Sum the components along the i^ \hat{i} direction: Fnet,i^=300.0+(−200.0)=100.0 F_{\text{net}, \hat{i}} = 300.0 + (-200.0) = 100.0
Sum the components along the j^ \hat{j} direction: Fnet,j^=500.0+(−800.0)=−300.0 F_{\text{net}, \hat{j}} = 500.0 + (-800.0) = -300.0
Thus, the net force in component form is: F⃗net=(100.0i^−300.0j^) \vec{F}_{\text{net}} = (100.0 \hat{i} - 300.0 \hat{j})

STEP 4

Calculate the magnitude of the net force using the Pythagorean theorem.
The magnitude of the net force ∣F⃗net∣ |\vec{F}_{\text{net}}| is given by: ∣F⃗net∣=(Fnet,i^)2+(Fnet,j^)2 |\vec{F}_{\text{net}}| = \sqrt{(F_{\text{net}, \hat{i}})^2 + (F_{\text{net}, \hat{j}})^2} ∣F⃗net∣=(100.0)2+(−300.0)2 |\vec{F}_{\text{net}}| = \sqrt{(100.0)^2 + (-300.0)^2} ∣F⃗net∣=10000+90000 |\vec{F}_{\text{net}}| = \sqrt{10000 + 90000} ∣F⃗net∣=100000 |\vec{F}_{\text{net}}| = \sqrt{100000} ∣F⃗net∣=316.23 |\vec{F}_{\text{net}}| = 316.23

STEP 5

Calculate the direction of the net force using the arctangent function.
The direction θ \theta of the net force is given by: θ=tan⁡−1(Fnet,j^Fnet,i^) \theta = \tan^{-1}\left(\frac{F_{\text{net}, \hat{j}}}{F_{\text{net}, \hat{i}}}\right) θ=tan⁡−1(−300.0100.0) \theta = \tan^{-1}\left(\frac{-300.0}{100.0}\right) θ=tan⁡−1(−3.0) \theta = \tan^{-1}(-3.0) θ≈−71.57∘ \theta \approx -71.57^\circ
The negative angle indicates the direction is below the positive i^ \hat{i} -axis.
The net force on the telephone pole in component form is F⃗net=(100.0i^−300.0j^) \vec{F}_{\text{net}} = (100.0 \hat{i} - 300.0 \hat{j}) , with a magnitude of 316.23 316.23 and a direction of approximately −71.57∘ -71.57^\circ .

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