solution - phy11 -1

using trigonometry

The force in the wire corresponds to the hypotenuse in a right-angled triangle. The angle is 40°. The vertical force of 5 N is adjacent to the angle of 40°. So, the cosine is used.

$${ 5 N = F_{wire} \cdot \cos{40°} }$$

So,

$${ F_{wire} = \frac{5 N}{\cos{40°}} \approx 6.53 N }$$


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