A force acting on a body (i.e., object) is represented mathematically by a
vector. A *force diagram, *which is also known as a
*free body diagram, *is a sketch in which all the force vectors acting
on an object are drawn with their initial points at the location of the
object.

A force **F** acting on a body (i.e., object) is
represented mathematically by a vector, in that it has a magnitude ||**F**|| which is applied in a given direction. The magnitude ||**F**|| is measured in *Newtons, *where 1 Newton is the force required to accelerate a 1 kg mass at
a rate of 1 meter per second per second (i.e, 1 *N* = 1 *kg*·*m*/sec^{2}
).

A *force diagram, *which is also known as a *free body diagram, *is a sketch in which all the force vectors acting on an object are drawn
with their initial points at the location of the object. The *net
force *acting on the body is the sum of all the forces in the force diagram.

EXAMPLE 8A 10kg block resting on a table is attached by a "massless" rope to a 5 kg block hanging off the end (with the rope running across a "frictionless" pulley).

Draw the free-body diagram and determine the net force if the 10kg block is experiencing a force of friction with a magnitude of 30 newtons.

Solution:To begin with, the force of gravityF_{gravity}is pulling the 10 kg block toward the earth, which via Newton's third law is countered by a force normal (i.e., perpendicular) to the table that resists the 10kgblocks fall. If the block is not moving vertically, thenF_{normal}andF_{gravity}must have the same magnitude so that they cancel each other out (i.e, no net force up or down).Using Newton's law

F=ma, we can show that the 5kg block is subject to a gravitational force ofSince ||

F_{5kg}= 5kg·0,-9.8 msec

^{2}= 0,-49 msec

^{2}F_{5kg}|| = 49N, the 5kg block applies a force ofto the 10kg block. Finally, the force of friction

F_{applied}= 49iNewtonsF_{friction}resists the horizontal motion of the block, so it is applied in the direction -i, orThus, the net force is

F_{friction}= -30iNewtons

F_{net}= F_{gravity}+F_{normal}+F_{applied}+F_{friction}= 0+49 i-30i= 19 i

In example 8, a net force of **F**_{net} = 19**i** means
that the 10kg block is accelerating to the right at a rate of 1.9 meters per
second per second. In contrast, had the friction had a magnitude of 49
Newtons, it would not be moving at all - i.e., it would be at rest with the
5kg block likewise at rest as it hangs suspended in midair.

If the net force acting on a body is 0, then the body is said to be at *equilibrium. *When it occurs, equilibrium often can be used to determine
unknown quantities in a free body diagram.

EXAMPLE 9A 50kg block is suspended by 2 identical cables of the same length forming 45^{°}angles with the horizontal.

If the block remains motionless over time, then what is the magnitude of the tensile force exerted by each of the cables on the block?

Solution:In this case, the free body diagram of the 50 kg mass is given by the figure below.The force of gravity is

If we let

F_{gravity}= 50kg· á 0,-9.8m/ sec^{2}ñ = á 0,-490NñFdenote the (unknown) magnitude of the force exerted by a cable, thenConsequently, the net force is

F_{1}

=

Fá cos(135^{° }) , sin(135^{° }) ñ =

- FÖ2

, FÖ2

F_{2}

=

Fá cos(45^{° }), sin(45^{° }) ñ =

FÖ2

, FÖ2

F_{net}= F_{gravity}+F_{2}+F_{1}

=

á 0,-490 ñ + FÖ2

, FÖ2

+

- FÖ2

, FÖ2

=

0,-490+ 2 FÖ2

Since

F_{net}= 0 (equilibrium), we have

2 FÖ2

= 490, F=490Ö2 2

= 346.48 N