Horizontal force and Tension

In summary: Fsin([alpha]).N = mgcos([alpha]) = 1.36(9.81)cos(23.6) = 12.35NF = 1.36(9.81)sin(23.6) = 5.78NIn summary, the magnitude of the horizontal force F holding a 3.87 kg block in equilibrium on an incline with a 23.6 degree angle is 16.6N. The tension in a wire produced by a 1.36 kg bird landing on it midway between two telephone poles, causing the wire to sag 0.118 m, is 1357N with a coefficient of friction of 0.261.
  • #1
Aerospace
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0
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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  • #2
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...)
 
  • #3
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
 
  • #4
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
 
  • #5
I am not sure if it attached the file earlier...hmmmm
 
  • #6
Last try
 

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  • #7
Resolve parallel to the inclined surface.

FCosA = mgSinA + umgCosA

where u = coefficient of friction, m = mass, g = gravity, A = slope angle.
 
  • #8
Condition of equilibrium:

Nsin([alpha])=F,
N=mgcos([alpha])+Fsin([alpha]),
mg=Ncos([alpha])
 

Attachments

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1. What is horizontal force?

Horizontal force refers to a force that is applied in a horizontal direction, parallel to the ground. This type of force can cause an object to move horizontally or change its direction.

2. How is horizontal force measured?

Horizontal force is typically measured in units of newtons (N) using a force meter or a spring scale. It can also be calculated by multiplying the mass of an object by its acceleration.

3. What is tension?

Tension is a force that is transmitted through a rope, cable, or other similar object when it is pulled tight. It acts in a direction that is parallel to the object and can cause it to stretch or deform.

4. How is tension related to horizontal force?

Tension and horizontal force are closely related, as tension can be created by applying a horizontal force to an object. For example, when pulling a wagon with a rope, the tension in the rope is caused by the horizontal force applied to it.

5. What are some real-life examples of horizontal force and tension?

Some common examples of horizontal force and tension include pushing a heavy object across a surface, pulling a door open, and swinging on a rope. In addition, bridges, cranes, and zip lines all rely on horizontal force and tension to function properly.

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