Maximum Horizontal Force?

Question

A sled weighing 61.0 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is 0.100. A penguin weighing 73.0 N rides on the sled, as in Fig. P4.76. If the coefficient of static friction between penguin and sled is 0.700, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

I'm a junior in an AP Physics class. I asked my intructor to explain this question, but he only confused me more....please help ASAP....it's due tonight....Best Answer points to first correct answers with a few explantions......thanks!

Answer 1

Consider forces acting on a sled and on a penguin.
Forces on penguin: friction with sled F1, gravity Mg = 73N and reaction from sled N1. Vertically forces are balanced so N1 = Mg = 73N. Friction with sled F1 provides horizontal acceleration a1 = F1/M. Because F1<=0.7N1=51.1N maximum acceleration that penguin could experience is a1 <= 0.7*N1/M = 0.7g = 7 m/s^2 (I assume g = 10)
Forces on sled: pulling force F, friction with snow F2, friction with penguin F1, gravity mg = 61N, penguin's weight Mg = 73N and reaction N2. Vertically forces are balanced so N2 = mg + Mg = 134N and kinetic friction is F2 = 0.1*N2 = 13.4N.
Sled would experience horizontal acceleration ma2 = F - F1 - F2 = F - 51.1 - 13.4 = F - 64.5. Then F = ma2 + 64.5
Penguin does not slide when a1 = a2, so maximum acceleration when penguin does not slide a2 = a1 = 0.7g, then ma2 = 0.7mg = 0.7*61 = 42.7N. Then maximum force F = 42.7 + 64.5 = 107.2N.
When F is larger then this value sled would accelerate faster than penguin and penguin would slide off from sled.
Check my calculations.

Answer 2

The coefficient of friction tells you the largest force that can develop to resist relative motion between two surfaces for a given normal force pushing the two surfaces together.

If the coefficient of friction is 0.1, then for every 1N normal force pushing the two surfaces together, the largest shear force that they can support is 0.1*1N=0.1N. There are two common coefficients of frictions. One is the STATIC coefficient of friction. The other is the KINETIC coefficient of friction. The static one decribes the shear force in order to get two surfaces, which are initially not moving with respect to each other, to start moving with respect to each other. The kinetic coefficient of friction describes the shear force for two surfaces that are already moving with respect to each other.

I can't see your figure P4.76 so I have no idea what it is showing. I'm going to assume that there is a sled on a flat horizontal ground covered in snow and is being pulled horizontally. There is a penguin sitting on top at the center of the sled.

In the simplest interpretation of your question, the weight of the sled and the kinetic coefficient of friction between it and the snow is irrelevant. You wanto to know the maximum amount of shear force possible between the sled and the penguin's butt. Anything larger than that and the penguin starts to slide on the sled. So in that case, you want the static coefficient of friction between the penguin and the sled, which you are given as 0.7. You need to multiple this by the normal force pushing the penguin and the sled together, which is just the weight of the penguin. So 0.700*73.0N=51.1N. That's the hardest you can pull on the sled without the penguin sliding.

The other details about the weight of the sled and the firction between it and the snow describes the forces that develop to pull the sled across the snow. It has nothing to do with whether the penguin stays on it.

Answer 3

Have you seen a trick where a table cloth being pulled out from under the plates and the plates remain on the table?

This is a similar problem.
You are applying force on the tablecloth and the tablecloth applies force on the plates...

The force on the sled, penguin and against the friction between the snow and the sled is

F=(m1 + m2)a + u1(m1 + m2)g
u1 - is th coefficient of friction between the snow and the sled

The force that holds our penguin on the surface of the sled is
f=u2(m2g)
If we exceed this force the penguin will 'fly'.

The acceleration that will make the penguin fly is
a2=f/m2 =u2(m2g)/m2=u2g

Lets revisit our old equation

F=(m1 + m2)a + u1(m1 + m2)g [this a is our a2 is it not?]
so
F=(m1+m2) (u2 g) + u1(ma1+m2)g or simply

F=(u2 + u1)(m1 + m2)g
F=(0.7 + 0.1)(61.0 + 73.0)=133 N

Answer 4

enable mass of sled be M1 & mass of penguin be M2 equation of action for sled F(max)-f(s)-f(ok)=M1a---A equation of action of penguin f(s)=M2a------B static friction= f(s)=49N; kinetic friction= f(ok)=6N a-- acceleration of sled & penguin;fixing A&B F(max)=104N