Question

A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) F⃗ 2=−200.0iˆ F⃗ 3=−800.0jˆ. (a) Find the force on the telephone pole in component form. (b) Find the magnitude and direction of this net force.

Answer 1

As per the given question,

Force on the pole $F_1=300.0\hat{i}+500.0\hat{j}$

$F⃗ _2=−200.0\hat{i}$

$F⃗ 3=−800.0\hat{j}$

a)

Net force on the pole $=F_1+F_2+F_3$

$=300.0\hat{i}+500.0\hat{j}−200.0\hat{i}−800.0\hat{j}=100\hat{i}-300\hat{j}$

Hence the component of the force on the pole is 100N along the x axis, and 300 along the y axis towards the ground.

b)

Net force on the pole =$|100\hat{i}-300\hat{j}|=\sqrt{100^2+300^2}=100\sqrt{10}N$

The direction of the net force $\tan\theta =\dfrac{F_y}{F_x}=\dfrac{-300}{100}=\dfrac{-3}{1}$

Hence, $\theta=71.56^\circ$ downwards.