Condition satisified when the body does not slide

• Eugen
In summary, the equation for static friction between m2 and m1 sledges is greater than the tension in the string, and so m2 does not slide.
Eugen

Homework Statement

A sledge of mass m1 is pulled horizontally with a force F. On the sledge there is a body of mass m2 that can slide on the horizontal platform of the sledge with the friction coefficient μ. Another sledge of mass m3 is tied with a horizontal string of the body m2. Between the sledges and the snow the friction is negligible. Find the condition that is satisfied when the
m2 body does not slide.

Homework Equations

fs = μN

T = [m2/(m1 + m2)]*F

The Attempt at a Solution

I assumed that μ is the coefficient of static friction. If the m2 body is not to slide, the static friction between this body and the m1 sledge must be bigger than the tension in the string.
The tension in the string should be:

T = [m3/(m1 + m2 + m3)] * F

and the static friction:

f = m2

m2gμ > [m3/(m1 + m2 + m3)] * F

This gives:

μ > F/m2g * [m3/(m1 + m2 + m3)]

but the book answers say that:

μ > F/m2g * [(m3 + m2)/(m1 + m2 + m3)]

What is wrong?

Eugen said:
If the m2 body is not to slide, the static friction between this body and the m1 sledge must be bigger than the tension in the string.
Draw a free body diagram for m2 and set up the second law for m2. This will give you the correct relation between the static friction force, the tension, and the acceleration.

The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force:

F' = m2 F/(m1 + m2 + m3)

The tension acts horizontally in the negative direction:

T = - m3F/(m1 + m2 + m3)

So, on m2 acts horizontally this force:

Fr = F(m2 - m3)/(m1 + m2 + m3)

which can be positive or negative. The friction should compensate for the resultant force, so it should be positive and equal in magnitude with the tension.

But, the solution says that
μ > F/m2g * [(m2 + m3)(m1 + m2 + m3)]

which in my opinion means that the friction should be bigger that the F' + T. I don't understand where this sum comes.

Last edited:
Eugen said:
The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force:

F' = m2 F/(m1 + m2 + m3)
Yes, that represents m2a. So it must equal the net force acting horizontally on m2. That is, it equals ∑Fx.

The tension acts horizontally in the negative direction:

T = - m3F/(m1 + m2 + m3)
Yes.

So, on m2 acts horizontally this force:

Fr = F(m2 - m3)/(m1 + m2 + m3)
There is a sign error here. Start with ∑Fx = m2ax. On the left side, add all the x-components of the forces acting on m2, taking into account the signs of the x-components. Then solve for Fr.

The friction should compensate for the resultant force, so it should be positive and equal in magnitude with the tension.
If the friction force is equal in magnitude to the tension force (but opposite in direction), would m2 have any acceleration?

Eugen
Oh. Now I get it. The force acting on m2 is the resultant of tension and friction, the friction being in the positive direction:

-m3F/(m1 + m2 + m3) + m2gμ = m2F/(m1 + m2 + m3)

Thank you.

1. What is meant by "condition satisfied when the body does not slide"?

In physics, the term "condition satisfied when the body does not slide" refers to the state in which an object or body is at rest or moving at a constant velocity without sliding or slipping. This means that the force acting on the body is balanced, resulting in no acceleration or change in motion.

2. What factors affect the condition satisfied when the body does not slide?

There are several factors that can affect the condition satisfied when the body does not slide, including the mass of the object, the surface it is resting on, and the forces acting upon it. Friction, gravity, and applied forces are some of the main factors that can impact this condition.

3. How is friction involved in the condition satisfied when the body does not slide?

Friction plays a crucial role in the condition satisfied when the body does not slide. It is the force that opposes the motion of an object and is necessary to prevent an object from sliding or slipping. In order for the condition to be satisfied, the frictional force must be equal and opposite to the applied forces acting on the object.

4. Can the condition satisfied when the body does not slide be achieved on any surface?

No, the condition satisfied when the body does not slide can only be achieved on surfaces that provide enough friction to counteract the applied forces. For example, a flat, rough surface will provide more friction than a smooth, slippery surface, making it easier to satisfy this condition.

5. How is the condition satisfied when the body does not slide related to Newton's First Law of Motion?

The condition satisfied when the body does not slide is directly related to Newton's First Law of Motion, also known as the law of inertia. This law states that an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity, unless acted upon by an external force. When the condition is satisfied, the object's motion remains constant due to the balance of forces acting on it, in accordance with Newton's First Law.

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