What Is the Maximum Horizontal Force Before the Penguin Slides Off the Sled?

AI Thread Summary
The discussion centers on calculating the maximum horizontal force that can be exerted on a sled with a penguin on it before the penguin slides off. The sled weighs 60 N, the penguin weighs 70 N, and the coefficients of kinetic and static friction are given. An initial calculation suggests a force of 111 N, but the book states the answer is 104 N. Participants emphasize the importance of including acceleration in the force diagrams and equations to accurately determine the maximum force. The conversation highlights the need for careful consideration of static friction and Newton's laws in solving the problem.
Nuha99
Messages
15
Reaction score
0

Homework Statement



A sled weighing 60 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is uk = 0.100. A penguin of 70 N rides on the sled. If the co. static friction between the penguin and the sled is us = 0.700 find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.


Homework Equations



fs = us R, fk = uk R.

The Attempt at a Solution



Please see the attached file for the force diagrams.

Forward force on the sled = F - fs - fk

In the absence of the static friction, penguin will slide backward with this force. But the penguin expereinces a force due to static friction fs.

F - fs - fk = fs

F = 2 fs + fk = 2 us R1 + uk R2 = 2 us m1g + uk (m1+m2)g = 111 N

Book answer is 104 N.

Can you check my force diagrams please?

Thanks.
 

Attachments

  • friction.JPG
    friction.JPG
    6.9 KB · Views: 418
Physics news on Phys.org
Nuha99 said:
A sled weighing 60 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is uk = 0.100. A penguin of 70 N rides on the sled. If the co. static friction between the penguin and the sled is us = 0.700 find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

Hi Nuha99! :smile:

I can't see your diagrams yet, but nor can I see anywhere in your equations where you've worked out the acceleration that the penguin can take.

Hint: start by applying Newton's second law to the penguin, and find that acceleration. :smile:
 
Attached file is not validated yet.

I did not find the acceleration of the sled. I found the net forward force on the sled. According to my understanding this net forward force is what will make the penguin to slide off.

Forward force on the sled = F - fs - fk

Note that fs is the static frictional force on penguin between penguin and the sled surface. By Newton's 3rd low, same but opposite force should act on the sled. This is the origin of fs in the above expression.

Now for the penguin to be stable, static frictional force = F - fs - fk

fs = F - fs - fk

Apparently some thing is not right. I solved this for F.
See my first post.

Thanks.
 
Nuha99 said:
I did not find the acceleration of the sled. I found the net forward force on the sled. According to my understanding this net forward force is what will make the penguin to slide off.

Hi Nuha99! :smile:

If there is acceleration (which there is),

then the force diagram must include accleration …

and so must the equations.
 
That does it. Thank you.
 
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}...
Back
Top