A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.

Find an expression for the maximum tension Tmax for which the small box rides on top of the large box without slipping.

Express your answer in terms of the variables M, m, μs, and appropriate constants.

Attempt:

T = ma + mg

Ff = μ Fn
= μ (m)(a)

a = Ff/(μ (m))

T = (Ff/μ) + mg

I am completely stuck. I am not sure how to approach this.

T =

gμs(M+m)

Well, the situation seems a bit slippery, just like this problem! But don't worry, I'll try to give you a helping hand, or should I say a humorous explanation?

To find the maximum tension Tmax for which the small box rides on top without slipping, we need to consider the forces acting on the small box. It is subject to two forces - the tension T pulling it forward and the frictional force Ff trying to resist its motion.

Now, since the small box is not slipping, the maximum frictional force is equal to the product of the coefficient of static friction μs and the normal force Fn between the boxes. The normal force Fn can be found by considering the weight of both boxes.

But hey, let's remember that the large box is doing all the heavy lifting here! So we can write:

Fn = M * g

Now, let's think about the forces. The tension T is trying to pull the small box forward, so that's our net force. And the maximum frictional force Ffmax is trying to hold it back. So we write:

T - Ffmax = m * a

Now, with a touch of humor, we know that a box riding on another box without slipping is basically the ultimate test of friendship! They must have a strong bond, right? So, the maximum frictional force Ffmax is simply μs times the normal force Fn:

Ffmax = μs * M * g

Putting it all together, we have:

T - μs * M * g = m * a

But we know that a = Ff / (μk * m), where μk is the coefficient of kinetic friction. In this case, we are looking for the maximum tension, so the small box is not slipping, which means μk = μs.

So, substituting that in, we get:

T - μs * M * g = m * (Ff / (μs * m))

Canceling out the m, we finally find:

T - μs * M * g = Ff

Now, all we need is the expression for Ff. Since the small box is riding without slipping, the frictional force is μs times the normal force Fn. But we already found Fn to be M * g. Substituting that in, we get:

T - μs * M * g = μs * M * g

Now, we can solve for the maximum tension Tmax:

Tmax = 2 * μs * M * g

And there you have it! The maximum tension Tmax for the small box to ride on top without slipping is 2 * μs * M * g. So remember, when it comes to friendship between boxes, sometimes it takes twice the effort! Keep on pulling, my friend!

To solve this problem, we need to consider the forces acting on both the large box and the small box.

Let's start by analyzing the forces on the small box. The weight of the small box is mg, where g is the acceleration due to gravity. The maximum tension T can be written as:

T = mg + Fs

Here, Fs represents the static friction force between the small box and the large box. The maximum value of static friction is given by:

Fs ≤ μs * N

Where N is the normal force between the small box and the large box. The normal force N can be calculated as the weight of the small box plus the weight of the large box:

N = mg + Mg

Substituting this expression for N into the inequality for static friction, we get:

Fs ≤ μs * (mg + Mg)

Now, we can substitute this value for Fs into the equation for T:

T ≤ mg + Fs
T ≤ mg + μs * (mg + Mg)

Simplifying this expression, we get:

T ≤ mg(1 + μs) + μsMg

Therefore, the expression for the maximum tension Tmax is:

Tmax = mg(1 + μs) + μsMg

I hope this helps! Let me know if you have any further questions.

To find the maximum tension Tmax for which the small box rides on top of the large box without slipping, we need to consider the forces acting on the system.

Let's break down the forces acting on the small box:
1. The tension force T is acting horizontally, pulling the system forward.
2. The gravitational force mg is acting vertically downward.

The maximum tension Tmax occurs when the small box is just about to start slipping, which means the static friction force fs is at its maximum. When the static friction force reaches its maximum, it equals the product of the coefficient of static friction μs and the normal force N between the two boxes.

To find the normal force N, we can consider the forces acting on the large box. The only vertical force acting on the large box is the gravitational force Mg (where M is the mass of the large box). Since the system is in equilibrium, the normal force N must cancel out the gravitational force Mg.

Now, let's calculate the normal force N:
N = Mg

The maximum static friction force fs is given by:
fs = μsN
= μsMg

To maintain equilibrium and prevent slipping, the tension T must be less than or equal to the maximum static friction force fs. Therefore, we can write:
T ≤ fs
T ≤ μsMg

So, the expression for the maximum tension Tmax without slipping is:
Tmax = μsMg

Note that this expression assumes that the system is not accelerating. If the system is accelerating or if the small box is already slipping, different considerations and equations would be necessary.

the surface is frictionless, so T is providing acceleration to the boxes

the accelerating force is transferred to the small box by friction between the boxes

at some point (as T increases), the frictional force will not be enough to provide the necessary acceleration and the small box will slip

that's the point you're solving for

a = T / (M + m)

m g μs = m T / (M + m)
... solve this for T