Question

1. your are out fishing and your hook catches a tire (but you dont know this!), the tire is pulling downwards with 30 n of force. you are only able to pull on it at an angle of 55° degrees. you can pull up on the fishing pole with 38 newtons of force. what will be the magnitude of the net force? (just put the number do not put the unit, and round to the hundredths position if needed)
2. a man is attempting to pull a wooden box that is currently at rest. the wooden box is resting on a concrete floor. the box has a weight of 112 n. the man pulls with a force of 30 n at an angle of 30 degrees. what is the magnitude of the net force in the horizontal direction? (just put the number do not put the unit, and round to the hundredths position if needed)

Explanation:

Step1: Resolve the pulling force into vertical and horizontal components for question 10

For the 38 - N pulling force on the fishing - pole, the vertical component \(F_{y}\) is given by \(F_{y}=38\sin(55^{\circ})\).
\[F_{y}=38\times\sin(55^{\circ})\approx38\times0.8192 = 31.13\]
The net force \(F_{net1}\) in the vertical direction is \(F_{net1}=F_{y}-30\).
\[F_{net1}=31.13 - 30=1.13\]

Step2: Resolve the pulling force into vertical and horizontal components for question 11

For the 30 - N pulling force on the wooden box, the horizontal component \(F_{x}\) is given by \(F_{x}=30\cos(30^{\circ})\).
\[F_{x}=30\times\cos(30^{\circ})=30\times\frac{\sqrt{3}}{2}\approx30\times0.8660 = 25.98\]

Answer:

1.13
25.98