Help, static equilibrium solving for maximum angle of incline

In summary, a machine in an ice factory is able to exert 3.3 * 10^2 N of force to pull large blocks of ice up a slope. The blocks each weigh 1.21 *10^4 N and assuming no friction, the maximum angle that the slope can make with the horizontal is approximately 1.563 degrees. This can be calculated by equating the component of gravity and the pulling force of the machine along the incline, and using the equation -(cos^-1(Ft/Fg)-90)=x to solve for the angle.
  • #1
Runaway
48
0

Homework Statement


A machine in an ice factory is capable of exerting 3.3 * 10^2 N of force to pull large blocks of ice up a slope. The blocks each weight 1.21 *10^4 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task.
Fg= 1.21*10^4 N
Ft= 3.3 * 10^2 N
Fn is perpendicular to Ft

Homework Equations


(in case the symbols my teacher uses are not standard)
Fg= weight
Ft= tension force
Fn= normal force
F()x= the x component of said force ie Ftx
F()y= the y component of said force ie Fty

The Attempt at a Solution


These are the only equations I can come up with using the given data, but no matter how I substitute them, I have yet to find a solvable equation.
Ftx= Ft/sin([tex]\theta[/tex])
Fty=Ft/cos([tex]\theta[/tex])
Fnx=Fn/sin(90-[tex]\theta[/tex])
Fny=Fn/cos(90-[tex]\theta[/tex])
Ft= sqrt(Ftx2 + Fty2)
Fn= sqrt(Fnx2 + Fny2)
Fty+Fny=Fg
Ftx=Fnx
but no matter how I substitute them, I have yet to find a solvable equation.
ex.
1.21*10^4 N=(3.3 * 10^2 N)/cos([tex]\theta[/tex])+(sqrt(((3.3 * 10^2 N)/sin([tex]\theta[/tex])2 + ((1.21*10^4 N)-(3.3 * 10^2 N)/cos([tex]\theta[/tex]))2))/cos(90-[tex]\theta[/tex])
 
Last edited:
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  • #2
when the object is on an incline(without friction and with the machine pulling it) 2 forces act on it along the incline... i.e: the component of gravity and the pulling of the machine.

So equating those two you can get your answer..

PS:
There is no need of so many equations you've written up there..
 
  • #3
Thanks, I was completely over complicating the problem,
Am I looking at it correctly now,because I keep getting an answer of less than 2 degrees, which doesn't sound right?
2jko3.jpg

Ft=cos(90-x)*Fg
-(cos^-1(Ft/Fg)-90)=x
-(cos^-1(3.3*10^2/1.21*10^4)-90)=1.563 degrees
 
  • #4
yes looks right...
 
  • #5
thanks for your help :)
 
  • #6
my pleasure..:smile:
 

Related to Help, static equilibrium solving for maximum angle of incline

1. What is static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in no net force and no acceleration. This means that the object is either at rest or moving at a constant velocity.

2. How do you solve for the maximum angle of incline in static equilibrium?

To solve for the maximum angle of incline, you need to use the equations for forces and moments in static equilibrium. These equations include the sum of all forces in the x and y directions, as well as the sum of all moments about a chosen point. By setting these equations equal to zero and solving for the unknown angle, you can find the maximum angle of incline at which the object will remain in static equilibrium.

3. What are the key factors that affect the maximum angle of incline in static equilibrium?

The maximum angle of incline in static equilibrium is affected by the weight of the object, the coefficient of friction between the object and the surface it is on, and the distribution of weight on the object. These factors determine the forces and moments acting on the object and must be taken into account when solving for the maximum angle of incline.

4. Can the maximum angle of incline be greater than 90 degrees?

No, the maximum angle of incline in static equilibrium cannot be greater than 90 degrees. This is because at a 90 degree angle, the force of gravity acting on the object is in the same direction as the normal force from the surface, resulting in no friction force and no net torque. Any angle greater than 90 degrees would result in a net torque and cause the object to slide or topple.

5. How can the concept of static equilibrium be applied in real-life situations?

The concept of static equilibrium can be applied in various real-life situations such as building structures, bridges, and other engineering projects. It is also important in understanding the stability and safety of objects and structures, such as furniture, ladders, and vehicles. In addition, the principles of static equilibrium are used in fields such as physics, mechanics, and aerospace engineering.

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