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

  • Thread starter Thread starter mace42
  • Start date Start date
  • Tags Tags
    Law Newton's law
AI Thread Summary
The discussion revolves around solving a physics problem involving a penguin on a sled, where a horizontal force is applied, and friction is present. Key points include determining the tension in the cord and the sled's acceleration while considering the forces acting on both the sled and the penguin. Participants emphasize the importance of using Newton's laws correctly, particularly in identifying which forces contribute to the net force in the x-direction. The confusion about including the weight of the sled and penguin in the calculations is addressed, suggesting that both weights influence the normal forces and, consequently, the friction. Ultimately, the derived equation for acceleration is a=(Fa-ug(Mp+Ms))/Ms, with a calculated acceleration of 1.6 m/s².
mace42
Messages
3
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
  • Like
Likes mace42
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.
 
sled acceleration.JPG

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

Attachments

  • sled acceleration.JPG
    sled acceleration.JPG
    45.3 KB · Views: 586
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top