How Do You Calculate Horizontal Force and Tension in Physics Problems?

AI Thread Summary
To calculate horizontal force and tension in physics problems, it's essential to resolve force vectors into perpendicular components and apply Newton's first law. For the block on the incline, the horizontal force F can be determined using the equation F = mgsinA/cosA, yielding a result of 16.6 N. In the case of the bird on the wire, the tension T can be calculated using the relationship TSinA = W/2, resulting in a tension of 1357 N when considering the sag and distance between the poles. A clear diagram can aid in visualizing these forces and their components. Understanding these principles is crucial for solving similar physics problems effectively.
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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 homework 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
 
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
 

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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])
 

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