Spring on incline with static equlibrium, disposition unkown. .

In summary, the maximum amount the mass (3 kg) can be pulled up the incline and released while still maintaining static equilibrium is 13.05 cm. This is determined by setting up the equation for the forces of gravity and the spring, which must be equal to the force of friction acting in the opposite direction. The equation is (m*g*sinθ)+(μs*m*g*cosθ)=Kx, where m is the mass, g is the acceleration due to gravity, θ is the angle of the incline, μs is the coefficient of static friction, and K is the spring constant. This equation yields a value of 13.05 cm, which is the maximum distance the mass can be displaced while remaining in
  • #1
grtdane17
6
0
A mass (m = 3 kg) is connected to a spring of constant k (k = 50 N/m), where the mass is connected at the top to a spring on a fixed incline. The coefficient of static friction between the mass and the incline is µ(subscript s) = 0.6 and the angle that the incline makes with the horizontal is theta = 20º. The mass is initially held in place so that the spring is at its natural length.

By what maximum amount can the mass be pulled up the incline and released and still have the mass in static equilibrium? Write your final answer as x(subscript s) = ___ centimeters.

Currently I have the equation set up for the force of gravity + force of the spring = force of friction, with 13.05 cm as my disposition. Is this correct and if not then why.
 
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  • #2
Draw a free body diagram.
Align axis with the slope.
Resolve gravity into components.

You need to sum the forces.
If there was no friction, you could do it right?

The net force on the mass without friction needs to be less than the friction force.
Will gravity and the force from the extended spring act in the same direction?

Show your derivation and working.
 
  • #3
My work is as follows

(m*g*sinθ)+(Kx)=(μs*m*g*cosθ)

10.055+50x=16.58

50x=6.525

x=13.05 cm

I set the force of gravity and the force of the spring in the same downward direction bc when the spring becomes stretched upwards it wants to return to equilibrium, which is downwards, and the static friction would point in the opposite direction countering those forces, so that it remains in equlibrium.
 
  • #4
Well, presumably the spring is hanging from a high point? So you are compressing the spring? I'd just check the wording so you are sure you are displacing the mass in the right direction. Apart from that, you've done what I'd do. Don't forget the fbd in your answer.
 
  • #5
Sorry the spring is connected at the bottom of the incline

spring.jpg
 
  • #6
grtdane17 said:
My work is as follows

(m*g*sinθ)+(Kx)=(μs*m*g*cosθ)

10.055+50x=16.58

50x=6.525

x=13.05 cm

I set the force of gravity and the force of the spring in the same downward direction bc when the spring becomes stretched upwards it wants to return to equilibrium, which is downwards, and the static friction would point in the opposite direction countering those forces, so that it remains in equlibrium.

Your calculations and assigned values look good - that should be the answer.
 
  • #7
Thanks so much, my last question is does it make sense that the same disposition of 13.05 cm when it is strectched and remains at static equilibrium would then also apply to when the spring is compressed and remains at static equlibrium. I believe the equation then should look like

(m*g*sinθ)=(Kx)+(μs*m*g*cosθ)

10.055=50X+16.58

-6.525=50x

X= -.1305

13.05 cm

or should friction be pointing downwards with gravity
(m*g*sinθ)+(μs*m*g*cosθ)=Kx
which would then give us 53.27 cm
 
  • #8
I concur with PeterO - pat yourself on the back.

[edit] posts passed each other.
check direction of friction
 
  • #9
grtdane17 said:
Thanks so much, my last question is does it make sense that the same disposition of 13.05 cm when it is strectched and remains at static equilibrium would then also apply to when the spring is compressed and remains at static equlibrium. I believe the equation then should look like

(m*g*sinθ)=(Kx)+(μs*m*g*cosθ)

10.055=50X+16.58

-6.525=50x

X= -.1305

13.05 cm

or should friction be pointing downwards with gravity
(m*g*sinθ)+(μs*m*g*cosθ)=Kx
which would then give us 53.27 cm

Yes friction is reversed - so the last bit is better [and presumably correctly calculated]
 

What is the concept of "Spring on incline with static equilibrium, disposition unknown"?

The concept of "Spring on incline with static equilibrium, disposition unknown" refers to a physics problem where a spring is attached to an inclined plane and is in a state of static equilibrium, but the direction of its displacement is unknown. It is a common problem in classical mechanics that requires the application of Newton's laws of motion.

What is static equilibrium?

Static equilibrium is a state in which an object is at rest or moving at a constant velocity with no net force acting on it. In other words, the object is balanced and the forces acting on it are equal in magnitude and opposite in direction.

How is the static equilibrium of the spring on incline determined?

The static equilibrium of the spring on incline can be determined by analyzing the forces acting on the system. These forces include the weight of the object, the normal force from the incline, and the force of the spring. By setting up and solving the equations for the forces in the x and y directions, the displacement and direction of the spring can be determined.

What factors affect the static equilibrium of the spring on incline?

The factors that affect the static equilibrium of the spring on incline include the mass of the object, the angle of the incline, the spring constant, and the initial displacement of the spring. A change in any of these factors can alter the equilibrium of the system and affect the direction of the spring's displacement.

How is this concept applied in real-life situations?

This concept is applied in many real-life situations, such as the suspension system of a car, the balance of a seesaw, or the stability of a structure on an inclined surface. It is also used in engineering and design to ensure that structures and objects are in a state of equilibrium to prevent tipping or collapsing.

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