Find the force on the cable, beam, and pole

In summary, the problem involves finding the force on a cable, beam, and pole that are supporting a mass of 18kg. The cable is at a 32 degree angle with the beam and attached to the pole. Newton's second law is not applicable as everything is stationary, so components or moments must be used. The acceleration due to gravity is 9.8m/s, but there must be another force acting on the object. The net force in both the x and y direction must be zero for equilibrium. The formula for this problem is \sum{F} = ma, but since it is 2D, vector must be used. A force diagram must be drawn and basic trigonometry can be used to solve the problem
  • #1
Joe91090
54
0

Homework Statement



a mass of 18kg is supported at the end of a horizontal beam hinged to a vertical pole, a cable at 32 degrees with the beam is attached to the pole to support the mass.

Find the force on the cable, beam, and pole.

Homework Equations




f=ma


The Attempt at a Solution



I don't know how to calculate force with an angle do you use cos or sin and what would you multiple it by ?
 
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  • #2
Did you draw out a picture of the situation?
 
  • #3
yes I am not sure how to post a picture though
 
  • #4
Joe91090 said:
I don't know how to calculate force with an angle do you use cos or sin and what would you multiple it by ?

HI Joe91090! :smile:

It's always cos (of the angle between the force and the particular direction) …

you only write sin if for some reason you're using 90º minus that angle

(for example, if you want the vertical component, and if θ is the angle from the horizontal, then the relevant angle is 90º - θ, so you use cos(90º - θ), which is sinθ :wink:)
 
  • #5
so is the equation F=m(cosθ)x a ??
 
  • #6
Joe91090 said:
so is the equation F=m(cosθ)x a ??

Is that supposed to be Newton's second law? :confused:

If so, forget it … everything is stationary, so a = 0.

Hint: do a force diagram, and then take components or moments :smile:
 
  • #7
in the problem is says the acceleration is 9.8m/s
 
  • #8
Well, yes that is what is feeling due to gravity but the particle isn't moving so that can't be the only acceleration it's experience, or more precisely, the only force.
 
  • #9
Joe91090 said:
in the problem is says the acceleration is 9.8m/s

Yes, that is g, the acceleration due to gravity.

Look, you need an equation relating all three forces …

what do you think that equation is going to be?
 
  • #10
would 15.56N be too small for the force on the cable. I did 18cos(32).
 
  • #11
I thought F=mg x cos(θ) but now I am thinking that's wrong
 
  • #12
Joe91090 said:
would 15.56N be too small for the force on the cable. I did 18cos(32).

Joe91090, I've no idea what you're doing! :cry:

Please show us your full calculations if you want us to check them :smile:
 
  • #13
ok so I am trying to solve for the force on the cable first. I really have no idea what formula to use all i know is that it is at 32 degrees from the horizontal with a 18 kg mass hanging from the end of it.
 
  • #14
You should probably end up with two equations 1 for the y direction and 1 for the x direction, the net force is zero, so identify the forces in each direction and set it equal to zero.
 
  • #15
the question seems so simple I am just looking for what formula or formulas i would use
 
  • #16
Most of these problems are all the same thing with small differences. Haven't you done a problem like this before or atleast seen the teacher go over a similar one?
 
  • #17
No i miss last class and my professor assigns homework online and it is due tomorrow
 
  • #18
http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed06.htm
 
Last edited by a moderator:
  • #19
First draw a force diagram, mark on all the forces, after that since its only three forces equilibrium so to use an equilibrium triangle is the easiest way to do this.
With the triangle, just mark on everything then solve it using basic trig.

I seen you made the mistake of thinking that it is only 18N going downward, but its not, you know the formula use it, :) F= ma

Since the object is not moving so the cable must provide an upward force of the same as the downward force.

With the force on the beam, just think about what i said above. :P

If you want to know the formula, its still F = ma, but since it's 2D, you must use vector.
So the formula is [tex]\sum{F}[/tex] = ma
 

Related to Find the force on the cable, beam, and pole

1. What does "finding the force on the cable, beam, and pole" mean?

Finding the force on the cable, beam, and pole refers to determining the amount of force or load that is being applied to these structures. This is important in engineering and construction to ensure that the structures can withstand the force and remain stable.

2. How is the force on the cable, beam, and pole calculated?

The force on the cable, beam, and pole is calculated using Newton's Second Law of Motion, which states that force is equal to mass times acceleration. The mass and acceleration of the structures must be known in order to calculate the force.

3. What factors affect the force on the cable, beam, and pole?

The force on the cable, beam, and pole can be affected by several factors such as the weight of the objects being supported, the angle at which the force is being applied, and external forces such as wind or earthquakes.

4. Why is it important to find the force on the cable, beam, and pole?

It is important to find the force on the cable, beam, and pole to ensure that the structures are strong enough to support the weight and forces they are subjected to. This information is crucial in the design and construction of buildings, bridges, and other structures.

5. How can the force on the cable, beam, and pole be measured?

The force on the cable, beam, and pole can be measured using various instruments such as strain gauges, load cells, and force transducers. These instruments can accurately measure the amount of force being applied to the structures.

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