# Inclined planes stuck on someting

• NUBIG10
In summary, the conversation discusses a problem involving an Atwood's machine with a sliding mass on an inclined surface with a coefficient of kinetic friction. The masses and initial height are given and the goal is to determine the direction of rotation of the pulley and the time it takes for the first mass to hit the ground. The problem is solved using force diagrams and the equation ƩF=ma. The final answer is found to be a=-.183 and one of the areas of doubt is the handling of friction in the equation.
NUBIG10

## Homework Statement

2) In the variant of Atwood’s machine shown below, a mass slides on an inclined surface 40θ above the horizontal with a coefficient of kinetic friction μk = 0.15. The masses M1 and M2 are 1 kg and 2 kg, respectively. Which way does the pulley rotate? If the lowest point of each mass is initially 2m above the ground (vertically), and the system has a very small but non‐zero initial speed, how long does it take the first mass to hit the ground?

F=MA

## The Attempt at a Solution

So here's how I went about the problem. I first got a force diagram going and found all the forces

F-parallel = 15.01N
F-Perpend=12.59N
F-gravity=19.6N
F-Tens=?
This is for Mass 2 on the incline

For mass 1 I got

F-gravity 9.8
F-Tens=?

So I have been trying to use the Ʃ all forces=ma for both equations, this gives me

M2= ƩF=ma=Ftens-Ffrict+Fparallel
ƩF=2kg*a=Ftens+(-.15*12.59N)+15.01N

M1=ƩF=ma=Ftens-Fgrav
ƩF=1kg*a=Ftens-1N

Then I put the above M1 equation into M2 and solve for acceleration then plugging in a into either equation to get Ftens. I already know the answer a=-.183 I just want to beat on my craft so when the test rolls around I can set up quick variable templates then plug and chug. Areas of doubt are,

-signs
-what to do about friction in the sum I already know its going anticlockwise so Friction is negative correct?
-Plugged it in many times and cannot get the known answer.

Thanks for any help and I will keep this post updated if I come across the answer before I receive help

one mistake the friction coefficient is multiplied to F parallel

for my final equation I get a= -m2g-Ffriction+Fparallel/Mass1+Mass2

Help

figured it out. Complete waste of time on here... Get an infraction quickly but no response.

I would approach this problem by first reviewing the given information and understanding the physical principles involved. In this case, we are dealing with an Atwood's machine, which consists of two masses connected by a pulley. The mass on the inclined plane will experience a component of its weight parallel to the plane, as well as a normal force perpendicular to the plane. We also need to consider the force of friction acting on the mass due to the coefficient of kinetic friction given.

To determine the direction of rotation of the pulley, we need to consider the net torque acting on it. Since the mass on the inclined plane is sliding down the plane, it will exert a force on the pulley in a counterclockwise direction. The mass hanging from the pulley will also exert a force in a counterclockwise direction. We then need to compare this with the force of friction acting on the mass on the inclined plane. If the force of friction is greater than the net counterclockwise force, the pulley will rotate clockwise.

To determine the time it takes for the first mass to hit the ground, we can use the equation ΣF=ma. We need to consider the forces acting on the first mass, which include its weight and the tension in the string. We also need to consider the acceleration of the mass, which will be negative due to it moving downward. We can then solve for the time it takes for the mass to reach the ground using the equation d=1/2at^2.

In summary, the key steps to solving this problem would be to carefully consider all the forces acting on the system and their direction, use the appropriate equations to determine the acceleration and tension in the string, and finally use the equation for motion to determine the time it takes for the first mass to hit the ground. It is important to carefully consider the signs and direction of the forces and to check your calculations to ensure they are consistent with the known answer.

## 1. What is an inclined plane?

An inclined plane is a simple machine that consists of a flat surface that is tilted at an angle. It is used to make it easier to move objects from a lower level to a higher level by reducing the amount of force needed.

## 2. How do inclined planes get stuck on something?

Inclined planes can get stuck on something when there is not enough force acting on the object to overcome the friction between the inclined plane and the object. This can happen if the object is too heavy or if the angle of the inclined plane is not steep enough.

## 3. How can you prevent an inclined plane from getting stuck?

To prevent an inclined plane from getting stuck, you can increase the force acting on the object by using a steeper angle or by using a pulley system to lift the object. You can also reduce the friction between the object and the inclined plane by using a smoother surface or adding lubricant.

## 4. What are some real-life examples of inclined planes getting stuck?

Examples of inclined planes getting stuck include a car getting stuck on a steep hill, a heavy object getting stuck on a ramp, or a person getting stuck while trying to climb a steep set of stairs.

## 5. How can you free an inclined plane that is stuck on something?

To free an inclined plane that is stuck on something, you can try using a stronger force to move the object, adjusting the angle of the inclined plane, or using a different type of simple machine, such as a lever or a pulley, to help move the object. In some cases, you may need to physically lift the object off of the inclined plane to free it.

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