**Example**

Set a book on a table (a mass of 1,000 *g*) and exert a pushing force on the book of 10.0 *N*. Now draw the object's free-body diagram and calculate the reaction of the table.

**Solution**

Start by drawing the free-body diagram that will show the three forces acting in the problem: the weight of the book on the table, the pushing force, and the reaction of the table (normal).

*m*_{book} = 1,000 g = 1.00 kg

*F*_{push} = 10.0 N

*N* = ?

The weight of the book is:

*W*_{book} = *m* · g = 1 kg · 9.8 m/s^{2} = 9.8 *N*

The total force on the table is then:

*F* = *W*_{book} + *F*_{push} = 19.8 *N*

And the reaction of the table according to the third law of motion is a force equal and opposite of *F*:

*N* = 19.8 *N* upward

As you can see, this force is larger than the weight of the object.

**Tension**

In the process of moving small or large objects, we often use tools. One of these tools is a simple string and is the base of more complex machines such as *pulleys*. A string in a pulley is used to transfer forces from the source of force to the object(s), to reduce the load on the string, and also to change the direction of motion. In the simplest case, it is considered that the string has no mass, hence no energy is needed to accelerate the string (there is no net force on the rope itself).

**Example**

Consider the following situation for which the freebody diagram needs to be determined.

**Solution**

The hanging body is acted upon only by its weight and the tension in the rope, which is the reaction of the force on the rope as it can be seen in Figure 5.8.

The object on the surface is acted upon by the tension in the rope and on the vertical by the weight and the normal force.

On the pulley, we have tension from both ends of the rope. (This is similar to a rubber band when you try to pull it down by putting your finger inside it and pulling on it; you can feel each part of the elastic band pulling on your finger.) Also, we have the reaction of the table to which the pulley is attached.

So, the free-body diagram treats the system of objects as independent objects, the connection between them being the tension force (the interaction force).

Kinetic Frictional Force

The magnitude f_{k}of the kinetic frictional force is proportional to the normal force N and to the coefficient of kinetic friction:

f_{k}= μ_{k}· N

where μ_{k}is the coefficient of kinetic friction and is a dimensionless quantity.

**Friction**

So, what happens then in terms of interaction between the object and the surface if we cannot neglect friction? At an atomic level, even the smoothest surface shows some roughness, and so if two such surfaces come into contact, there is a resistance to motion, which we call *friction*.

The forces of friction are classified in the following ways: (1) *static* frictional force can have a value of zero up to a maximum frictional force, *f*_{s}^{MAX}, and has a value dependent on the applied force, and (2) *kinetic* frictional force, which acts when the object is in motion and is directly opposite to the direction of motion.

**Example**

Consider a hockey puck of 500 g mass sliding on ice. If the puck's normal force is equal only to the weight of the object, what are the frictional force and the acceleration of the object if the coefficient of kinetic friction is 0.3? In the horizontal plane, consider only the frictional force.

**Solution**

In order to find the acceleration, construct a free-body diagram and calculate the frictional force. We know:

*m* = 500 g = 0.5 kg

μ_{k} = 0.3

W = *N*

Then we calculate the normal force (see Figure 5.11):

*N = m · g = 0.5 · 9.8 = 4.9 N*

*f _{k}* = μ

_{k}·

*N*= 0.3 · 4.9 = 1.5

*N*

*f _{k} = 1.5 N*

And because there is no other interaction, the second law of mechanics will give us the acceleration of motion. But before calculating acceleration, we determine that forces in the direction of motion will be considered positive and forces opposite to the motion will be negative.

∑*F* = *m* · *a*

∑*F* = *f*_{k} = *m* · *a*

*f*_{k} = *m* · *a*

Because the motion is in one direction, we can neglect the vector sign and write:

–μ_{k} · *N* = *m* · *a*

*a* = –μ_{k}· N/m = –μ_{k} · *m* · g/m =

–μ_{k}· g = –3 m/s^{2}

a = –3m/s^{2}

This value of the acceleration tells us that the object is slowing down.

The problem can also be solved numerically using the value of *f _{k}* = 1.5 N and applying the second law of motion:

a = –*f _{k}*/m = –1.5/0.5 kg = –3m/s

^{2}

**Static Equilibrium**

An object is in equilibrium when it has zero acceleration. The condition of equilibrium can be expressed as:

∑*F _{x}* = 0

∑*F _{y}* = 0

These conditions are equivalent to the fact that all forces acting on the object must balance. Then, in the nonequilibrium state, acceleration is nonzero:

∑*F _{x} = m a_{x}*

∑*F _{y} = m a_{y}*

### Ask a Question

Have questions about this article or topic? Ask### Related Questions

See More Questions### Popular Articles

- Kindergarten Sight Words List
- First Grade Sight Words List
- 10 Fun Activities for Children with Autism
- Grammar Lesson: Complete and Simple Predicates
- Definitions of Social Studies
- Child Development Theories
- Signs Your Child Might Have Asperger's Syndrome
- Social Cognitive Theory
- How to Practice Preschool Letter and Name Writing
- Theories of Learning