5.3 Newton's second law  (Page 5/8)

Force on a soccer ball

A 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by $\overrightarrow{a}=3.00\widehat{i}+7.00\widehat{j}{\text{m/s}}^2.$ Find (a) the resultant force acting on the ball and (b) the magnitude and direction of the resultant force.

Strategy

The vectors in $\widehat{i}$ and $\widehat{j}$ format, which indicate force direction along the x -axis and the y -axis, respectively, are involved, so we apply Newton’s second law in vector form.

Solution

1. We apply Newton’s second law:
   
   ${\overrightarrow{F}}_{\text{net}}=m\overrightarrow{a}=\left(0.400\text{kg}\right)\left(3.00\widehat{i}+7.00\widehat{j}{\text{m/s}}^2\right)=1.20\widehat{i}+2.80\widehat{j}\text{N}.$
2. Magnitude and direction are found using the components of ${\overrightarrow{F}}_{\text{net}}$ :
   
   $F_{\text{net}}=\sqrt{{(1.20\text{N})}^2+{(2.80\text{N})}^2}=3.05\text{N}\text{and}\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{2.80}{1.20}\right)=66.8\text{°}.$

Significance

We must remember that Newton’s second law is a vector equation. In (a), we are multiplying a vector by a scalar to determine the net force in vector form. While the vector form gives a compact representation of the force vector, it does not tell us how “big” it is, or where it goes, in intuitive terms. In (b), we are determining the actual size (magnitude) of this force and the direction in which it travels.

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Mass of a car

Find the mass of a car if a net force of $\mathrm{-600.0}\widehat{j}\text{N}$ produces an acceleration of $\mathrm{-0.2}\widehat{j}{\text{m/s}}^2$ .

Strategy

Vector division is not defined, so $m={\overrightarrow{F}}_{\text{net}}\text{/}\overrightarrow{a}$ cannot be performed. However, mass m is a scalar, so we can use the scalar form of Newton’s second law, $m=F_{\text{net}}\text{/}a$ .

Solution

We use $m=F_{\text{net}}\text{/}a$ and substitute the magnitudes of the two vectors: $F_{\text{net}}=600.0\text{N}$ and $a=0.2{\text{m/s}}^2.$ Therefore,

Significance

Force and acceleration were given in the $\widehat{i}$ and $\widehat{j}$ format, but the answer, mass m , is a scalar and thus is not given in $\widehat{i}$ and $\widehat{j}$ form.

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Several forces on a particle

A particle of mass $m=4.0\text{kg}$ is acted upon by four forces of magnitudes. $F_1=10.0\text{N},F_2=40.0\text{N},F_3=5.0\text{N},\text{and}{F}_4=2.0\text{N}$ , with the directions as shown in the free-body diagram in [link] . What is the acceleration of the particle?

Strategy

Because this is a two-dimensional problem, we must use a free-body diagram. First, ${\overrightarrow{F}}_1$ must be resolved into x - and y -components. We can then apply the second law in each direction.

Solution

We draw a free-body diagram as shown in [link] . Now we apply Newton’s second law. We consider all vectors resolved into x - and y -components:

Thus, the net acceleration is

which is a vector of magnitude $8.4{\text{m/s}}^2$ directed at $276\text{°}$ to the positive x -axis.

Significance

Numerous examples in everyday life can be found that involve three or more forces acting on a single object, such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We can see that the solution of this example is just an extension of what we have already done.

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