What is the Tension and Acceleration in Newton's Law of a Sled Problem?

In summary, the problem asks for the tension in a string that is attached to a sled and is pulling on a penguin. The string has a coefficient of friction of .2 and the sled is trying to move forward at an acceleration of 1.6 m/s/s.
  • #1
mace42
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Homework Statement


This is a problem from a post back in 2006. I have a fairly good understanding of how to do it; however, there is one part about the answer that I cannot figure out at all. (See the relevant equations)

Problem: A 5kg penguin sits on a 10kg sled. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow as well as that between the sled and the penguin is .2 b) Determine the tension in the cord and c) the acceleration of the sled if you try to solve for the tension using F=ma, do you could both the weights as the normal force? or do you only count the penguin's weight? Thanks

Reference https://www.physicsforums.com/threads/Newtons-law-problem-got-the-first-half.141064/

Homework Equations


On the post it said that the net force in the x direction of the sled was as follows:

Sum of all forces sub x = ma sub x
Fg + F sub f + (-Fx) = ma sub x

I do not understand why Fg was included as a force acting in the x direction. Is this wrong or am I missing something.

This is the equation I used:

Sum of all forces sub x = ma
F sub f + (-Fx) = ma

Any insight would be greatly appreciated!
 
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  • #2
Confusion arises by just trying to do the problem by looking at the equation for the x-motion. Do the problem completely. You don't have to guess, or be told by anyone, whether you should count both weights or only one. It will all come out as part of the solution.

So the forces on the sled are:
1. Pulling force 45 N
2. Friction from ground
3. Friction from penguin
4. Normal force from ground
5. Normal force from penguin
6. Weight of sled

The last three are in the y-direction, but they decide the friction forces, so let them be there. Draw the force diagram

Forces on the penguin are:
7. Tension from the string
8. Friction from sled
9. Normal force from sled
10. Weight of penguin

Draw the force diagram for this separately

Now use Newton's second law on each object, and then take components. Also use Newton's third law, which gives Force # 3 = #8; #5 = #9. Use this information and the answer will roll out automatically.
 
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  • #3
Chandra Prayaga said:
Confusion arises by just trying to do the problem by looking at the equation for the x-motion. Do the problem completely. You don't have to guess, or be told by anyone, whether you should count both weights or only one. It will all come out as part of the solution.

So the forces on the sled are:
1. Pulling force 45 N
2. Friction from ground
3. Friction from penguin
4. Normal force from ground
5. Normal force from penguin
6. Weight of sled

The last three are in the y-direction, but they decide the friction forces, so let them be there. Draw the force diagram

Forces on the penguin are:
7. Tension from the string
8. Friction from sled
9. Normal force from sled
10. Weight of penguin

Draw the force diagram for this separately

Now use Newton's second law on each object, and then take components. Also use Newton's third law, which gives Force # 3 = #8; #5 = #9. Use this information and the answer will roll out automatically.

Thank you so much for the help! Ultimately I got a=(Fa-ug(Mp+Ms))/Ms as an equation for acceleration and 1.6 m/s/s for an answer. If that isn't correct I can list my work.
 
  • #4
sled acceleration.JPG

I just discovered the upload button haha, here's my work.
 

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FAQ: What is the Tension and Acceleration in Newton's Law of a Sled Problem?

What is Newton's Law of a sled?

Newton's Law of a sled, also known as Newton's Second Law, states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that the greater the force applied to a sled, the greater its acceleration will be, and the heavier the sled, the slower it will accelerate.

How is Newton's Law of a sled applied in real life?

In real life, Newton's Law of a sled can be seen in various situations, such as when a person pushes a sled down a hill. The force of the push (net force) will determine how quickly the sled accelerates down the hill.

Is Newton's Law of a sled always applicable?

Newton's Law of a sled is applicable in most situations, but there are some exceptions. It does not hold true for objects moving at speeds close to the speed of light or in extreme situations such as in outer space with no external forces acting on the object.

How does friction affect Newton's Law of a sled?

Friction is a force that opposes motion, so it can affect the acceleration of a sled. The more friction there is, the more force is needed to overcome it and accelerate the sled. In some cases, friction can even cause the sled to decelerate instead of accelerate.

Can Newton's Law of a sled be used to calculate the motion of a sled?

Yes, Newton's Law of a sled can be used to calculate the motion of a sled. By knowing the mass of the sled, the net force acting on it, and the acceleration, we can use the formula F=ma to calculate the sled's motion and predict its future position.

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