## Horizontal force and Tension

Q1. A 3.87 kg block is held in equilibrium on an incline by the horizontal force F. The incline makes a 23.6 degree angle with the horizontal. Determine the magnitude of F. Answer in units of N.

Q2. The distance between two telephone poles is 48 m. When a 1.36 kg bird lands on the telephone wire midway between the poles, the wire sags 0.118 m. how much tension in the wire does the bird produce? Ignore the weight of the wire. Answer in units of N.

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We haven't covered these topic yet in my class, and there's not much in the book that is helping me since the hw assignment is based on a different book that the prof used last year. If you guys could give me some directions as to how to go about solving these questions - if I get these, I'm sure I'll get most of the Qs - if not all.

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 1. Are you familiar with the idea of resolving force vectors into perpendicular components? If so, do it and apply Newton's first law in different directions. If not, ask. It is vital. 2. This is similar too. Draw a good diagram and you'll see. (Assume the cable itself to lie in straight lines...)
 I agree with FZ+. But for your answers: 1) mgsinA=FcosA 3.87(9.81)tan(23.6) = F F = 16.6N 2) TSinA = W/2 A = atan(0.118/24) T = 1357N

## Horizontal force and Tension

Oops...I forgot to mention that the coefficient is 0.261
Ah...I'll just attach the question lol...The figures will help, am sure.
I was thinking that maybe I could use Fx = macosA but then I don't know the acceleration...or maybe mg? Gravity being 9.8 m/s^2

 I am not sure if it attached the file earlier...hmmmm
 Last try Attached Thumbnails
 Resolve parallel to the inclined surface. FCosA = mgSinA + umgCosA where u = coefficient of friction, m = mass, g = gravity, A = slope angle.
 Condition of equilibrium: Nsin([alpha])=F, N=mgcos([alpha])+Fsin([alpha]), mg=Ncos([alpha]) Attached Thumbnails