Question

Three forces of magnitudes F1=4.0N, F2=6.0N, and F3=8.0N are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (Figure 1) In this problem, you will determine the resultant (net) force by combining the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

Part A: Calculate the magnitude of the resultant force F? r=F? 1+F? 2+F? 3 acting on the block.

Express the magnitude of the resultant force in newtons to two significant figures.

Part B: What angle does F? r make with the positive x axis?

Express your answer in degrees to two significant figures.

Part C: What is the magnitude of the block's acceleration vector,

a??

Part D: What is the direction of a? ? In other words, what angle does this vector make with respect to the positive x axis?

Express your answer in degrees to two significant figures.

Part E: How far (in meters) will the block move in 5.0 s? Recall that it starts from rest.

Express the distance d in meters to two significant figures.

Answer 1

Concepts and reason

The concepts used to solve this problem are the component of vector in horizontal and vertical direction, addition of horizontal and vertical component separately, Newton’s second law of motion, and kinematic equation of motion.

The magnitude of the force and the angle can be calculated using the horizontal and the vertical component of the force.

The Newton’s Second law can be defined as the product of the force and the acceleration. The acceleration can be calculated using the Second law of Newton.

Fundamentals

For any vector ${\rm{\vec A}}$ can be express as,

${\bf{\vec A}} = {a_x}\hat x + {a_y}\hat y$

Here, ${a_x}$ is x-component and ${a_y}$ is y-component of ${\bf{\vec A}}$.

Then, the magnitude of the resultant vector is,

$\left| {\rm{A}} \right| = \sqrt {{{\left( {{a_x}} \right)}^2} + {{\left( {{a_y}} \right)}^2}}$

And the direction angle $\theta$ is,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{a_y}}}{{{a_x}}}} \right)$

The vector ${\rm{\vec A}}$ can write in form of magnitude and direction.

${\bf{\vec A}} = \left| {\rm{A}} \right|{\rm{cos}}\theta \hat x + \left| {\rm{A}} \right|\sin \theta \hat y$

Here, $\left| {\rm{A}} \right|$ is the magnitude of vector ${\bf{\vec A}}$ and $\theta$ is angle from $x$-axis.

According to Newton’s second law of motion the acceleration of any object is directly proportional to the magnitude of force applied and inversely proportional to the mass of the object.

In mathematical form,

$F = ma$

Here, $F$ is force applied, $m$ is mass and $a$ is acceleration.

The expression of kinematic motion equation is,

$S = ut + \frac{1}{2}a{t^2}$

Here, $u$ is the initial velocity, $a$ is the acceleration, $S$ is the distance covered by the object and $t$ is the time taken.

Write all the force in vector form.

The general vector equation in rectangular co-ordinate form is,

${\bf{\vec F}} = F\cos \theta {\rm{ \hat x}} + F\sin \theta {\rm{ \hat y}}$

$\begin{array}{c}\\{{\vec F}_1} = 4.0\cos \left( {25^\circ } \right)\hat x + 4.0\sin \left( {25^\circ } \right)\hat y\\\\ = 3.63\hat x + 1.69\hat y\\\end{array}$

And,

$\begin{array}{c}\\{{\vec F}_2} = 6.0\cos \left( {325^\circ } \right)\hat x + 6.0\sin \left( {325^\circ } \right)\hat y\\\\ = 4.91\hat x - 3.44\hat y\\\end{array}$

And,

$\begin{array}{c}\\{{\vec F}_3} = 8.0\cos \left( {180^\circ } \right)\hat x + 8.0\sin \left( {180^\circ } \right)\hat y\\\\ = - 8.0\hat x + 0\hat y\\\end{array}$

Now the resultant of all three forces is,

${\vec F_R} = {\vec F_1} + {\vec F_2} + {\vec F_3}$

Substitute $3.63\hat x + 1.69\hat y$ for ${\vec F_1}$, $4.91\hat x - 3.44\hat y$ for ${\vec F_2}$ and $- 8.0\hat x + 0\hat y$ for ${\vec F_3}$.

$\begin{array}{c}\\{{\vec F}_R} = \left( {3.63\hat x + 1.69\hat y} \right) + \left( {4.91\hat x - 3.44\hat y} \right) + \left( { - 8.0\hat x + 0\hat y} \right)\\\\ = \left( {3.63 + 4.91 - 8.0} \right)\hat x + \left( {1.69 - 3.44 + 0} \right)\hat y\\\\ = 0.54\hat x - 1.75\hat y\\\end{array}$

The magnitude of the resultant force $\left| {\vec F} \right|$ is,

$\left| {\vec F} \right| = \sqrt {{{\left( {{a_x}} \right)}^2} + {{\left( {{a_y}} \right)}^2}}$

Substitute 0.54 for ${a_x}$ and $- 1.75$ for ${a_y}$.

$\begin{array}{c}\\\left| {\vec F} \right| = \sqrt {{{\left( {0.54} \right)}^2} + {{\left( { - 1.75} \right)}^2}} \\\\ = 1.8{\rm{ N}}\\\end{array}$

Calculate the angle of resultant vector from x-axis.

The resultant vector is,

${\vec F_R} = 0.54\hat x - 1.75\hat y$

If the angle $\theta$ is angle from x-axis then,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{a_y}}}{{{a_x}}}} \right)$

Here ${a_y}$ is y component of resultant vector and ${a_x}$ is x component of resultant vector.

Substitute $- 1.75$ for ${a_y}$ and $0.54$ for ${a_x}$.

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 1.75}}{{0.54}}} \right)\\\\ = - 72.85^\circ \\\end{array}$

Now this angle from the positive x-axis is,

$\begin{array}{c}\\\theta = 360^\circ - 72.85\\\\ = 287^\circ \\\\ \approx 290^\circ \\\end{array}$

Calculate the magnitude of block acceleration.

The expression of force is,

$F = ma$

Here, $F$ is force applied, $m$ is mass and $a$ is acceleration.

Rearrange the equation $F = ma$ for $a = \frac{F}{m}$.

$a = \frac{F}{m}$

Substitute $1.8{\rm{ N}}$for $F$ and $2.0{\rm{ kg}}$ for $m$.

$\begin{array}{c}\\a = \frac{{1.8{\rm{ N}}}}{{2.0{\rm{ kg}}}}\\\\ = 0.9{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Calculate the direction of acceleration.

The direction of acceleration directly depends on the direction of force, it means the direction of acceleration is equal to the direction of acceleration.

So, the direction of acceleration is,

$\theta = 290^\circ {\rm{ }}$

Calculate the distance cover by block.

The expression of kinematic equation of motion is,

$S = ut + \frac{1}{2}a{t^2}$

Here, $u$ is the initial velocity, $a$ is the acceleration, $S$ is the distance covered by the object and $t$ is the time taken.

Substitute $0{\rm{ m/s}}$ for $u$, $0.9{\rm{ m/}}{{\rm{s}}^2}$ for $a$, and $5.0{\rm{ s}}$ for $t$.

$\begin{array}{c}\\S = \left( {0{\rm{ m/s}}} \right)\left( {5.0{\rm{ s}}} \right) + \frac{1}{2}\left( {0.9{\rm{ m/}}{{\rm{s}}^2}} \right){\left( {5.0{\rm{ s}}} \right)^2}\\\\ = 0 + \frac{1}{2}\left( {0.9{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {25.0{\rm{ }}{{\rm{s}}^2}} \right)\\\\ = 11.25{\rm{ m}}\\\end{array}$

Ans: Part A

The magnitude of resultant force is$1.8{\rm{ N}}$.