Chapter 1: Q6CP (page 14)

If ${\vec{F}}_{1} = <300, 0, - 200>$ and ${\vec{F}}_{2} = <- 150, 300, 0>$ , calculate the following quantities and make the requested comparisons: (a) ${\vec{F}}_{1} + {\vec{F}}_{2}$ (b) $|{\vec{F}}_{1} + {\vec{F}}_{2}|$ (c) $|{\vec{F}}_{1}| + |{\vec{F}}_{2}|$ (d) Is $|{\vec{F}}_{1} + {\vec{F}}_{2}| = |{\vec{F}}_{1}| + |{\vec{F}}_{2}|$ (e) ${\vec{F}}_{1} - {\vec{F}}_{2}$ (f) $|{\vec{F}}_{1} - {\vec{F}}_{2}|$ (g) $|{\vec{F}}_{1}| - |{\vec{F}}_{2}|$ (h) Is $|{\vec{F}}_{1} - {\vec{F}}_{2}| = |{\vec{F}}_{1}| - |{\vec{F}}_{2}|$ ?

Short Answer

Expert verified

Answer of each part is:

(a) ${\vec{F}}_{1} + {\vec{F}}_{2} = <450, - 300, - 200>$

(b) $|{\vec{F}}_{1} + {\vec{F}}_{2}| = 576.6$

(d) Not equal

(e) ${\vec{F}}_{1} - {\vec{F}}_{2} = <150, 300, - 200>$

(f) $|{\vec{F}}_{1} - {\vec{F}}_{2}| = 390.5$

(g) $|{\vec{F}}_{1}| - |{\vec{F}}_{2}| = 25.1$

(h) Not equal

Step by step solution

Basic rule of addition or subtraction of vectors

To add or subtract two vectors, add or subtract the corresponding components. This process is just like adding two constant numbers. Only the parts having same type of components need to be added or subtracted.

Finding sum of given vectors

(a)

Add the corresponding components of both the force vectors to obtain the sum of given vectors.

${\vec{F}}_{1} + {\vec{F}}_{2} = <300 + 150, 0 - 300, - 200 + 0>$

$= <450, - 300, - 200>$

Therefore, the vector forrole="math" localid="1653555770233" ${\vec{F}}_{1} + {\vec{F}}_{2}$ is $<450, - 300, - 200>$ .

Finding magnitude of sum of given vectors

(b)

Take the square root of sum of square of each component of the vector ${\vec{F}}_{1} + {\vec{F}}_{2}$ to obtain its magnitude.

$|{\vec{F}}_{1} + {\vec{F}}_{2}| = \sqrt{{(450)}^{2} + (- 300) 2 + {(- 200)}^{2}}$

$= \sqrt{202500 + 90000 + 40000}$

$= \sqrt{332500}$

= 576.6

Therefore, the value of $|{\vec{F}}_{1} + {\vec{F}}_{2}|$ is 576.6 .

Finding the value of F→1+F→2

(c)

Take the square root of sum of square of each component of the vector ${\vec{F}}_{1}$ to obtain its magnitude.

$|{\vec{F}}_{1}| = \sqrt{{(300)}^{2} + 0^{2} + {(- 200)}^{2}}$

$= \sqrt{90000 + 0 + 40000}$

$= \sqrt{130000}$

= 360.5

Take the square root of sum of square of each component of the vector ${\vec{F}}_{2}$ to obtain its magnitude.

$|{\vec{F}}_{2}| = \sqrt{{(150)}^{2} + {(300)}^{2} + 0^{2}}$

$= \sqrt{22500 + 90000 + 0}$

$= \sqrt{112500}$

= 335.4

Take the sum of obtained magnitudes.

$|{\vec{F}}_{1}| + |{\vec{F}}_{2}| = 360.5 + 335.4$

= 695.9

Therefore, the value of $|{\vec{F}}_{1}| + |{\vec{F}}_{2}|$ is 695.9 .

Analyzing F→1+F→2=F→1+F→2

(d)

From part (b), $|{\vec{F}}_{1} + {\vec{F}}_{2}| = 576.6$ and from part (c), $|{\vec{F}}_{1}| + |{\vec{F}}_{2}| = 695.9$ .

Therefore, the value of $|{\vec{F}}_{1} + {\vec{F}}_{2}|$ is not equal to the value of $|{\vec{F}}_{1}| + |{\vec{F}}_{2}|$ .

Finding the difference of given vectors

(e)

Subtract the corresponding components of both the force vectors to obtain the difference of given vectors.

${\vec{F}}_{1} - {\vec{F}}_{2} = <300 - 150, 0 + 300 . - 200 + 0>$

$= <150, 300 - 200>$

Therefore, the vector for ${\vec{F}}_{1} - {\vec{F}}_{2}$ is $<150, 300 - 200>$ .

Finding magnitude of difference of given vectors

(f)

Take the square root of sum of square of each component of the vector ${\vec{F}}_{1} - {\vec{F}}_{2}$ to obtain its magnitude.

role="math" localid="1653555991602" $|{\vec{F}}_{1} - {\vec{F}}_{2}| = \sqrt{{(150)}^{2} + {(300)}^{2} + {(- 200)}^{2}}$

$= \sqrt{22500 + 90000 + 40000}$

$= \sqrt{152500}$

= 390.5

Therefore, the value of $|{\vec{F}}_{1} - {\vec{F}}_{2}|$ is 390.5 .

Finding the value of F→1-F→2

(g)

From part (c), $|{\vec{F}}_{1}| = 360.5$ and $|{\vec{F}}_{2}| = 335.4$ .

Take the difference of obtained magnitudes.

$|{\vec{F}}_{1}| - |{\vec{F}}_{2}| = 360.5 - 355.4$

= 25.1

Therefore, the value $|{\vec{F}}_{1}| - |{\vec{F}}_{2}|$ of is 25.1 .

Analyzing F→1-F→2=F→1-F→2

(h)

From part (f), $|{\vec{F}}_{1} - {\vec{F}}_{2}| = 390.5$ and from part (g), $|{\vec{F}}_{1}| - |{\vec{F}}_{2}| = 25.1$ .

Therefore, the value of $|{\vec{F}}_{1} - {\vec{F}}_{2}|$ is not equal to the value of $|{\vec{F}}_{1}| - |{\vec{F}}_{2}|$ .

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Location	t(s)	Position ( m)
A	0.0	(0,0,0)
B	1.0	(22.3,26.1,0)
C	2.0	(40.1,38.1,0)

Short Answer

Step by step solution

Basic rule of addition or subtraction of vectors

Finding sum of given vectors

Finding magnitude of sum of given vectors

Finding the value of F→1+F→2

Analyzing F→1+F→2=F→1+F→2

Finding the difference of given vectors

Finding magnitude of difference of given vectors

Finding the value of F→1-F→2

Analyzing F→1-F→2=F→1-F→2

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If F→1=300,0,-200and F→2=-150,300,0, calculate the following quantities and make the requested comparisons: (a) F→1+F→2(b) F→1+F→2(c) F→1+F→2(d) Is F→1+F→2=F→1+F→2(e) F→1-F→2(f) F→1-F→2(g) F→1-F→2(h) IsF→1-F→2=F→1-F→2?

Short Answer

Step by step solution

Basic rule of addition or subtraction of vectors

Finding sum of given vectors

Finding magnitude of sum of given vectors

Finding the value of F→1+F→2

Analyzing F→1+F→2=F→1+F→2

Finding the difference of given vectors

Finding magnitude of difference of given vectors

Finding the value of F→1-F→2

Analyzing F→1-F→2=F→1-F→2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Kinematics Physics

Classical Mechanics

Thermodynamics

Electricity

Nuclear Physics

Study anywhere. Anytime. Across all devices.