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Animated Solution for Mathematics - Vector Algebra: The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be $19 n$ . The value of P is

Q: Wait, why does reversing the vector change the angle to $180^\circ - \alpha$?

Imagine two vectors $\vec{P}$ and $\vec{Q}$ with an angle $\alpha$ between them. If you reverse $\vec{Q}$ to become $-\vec{Q}$, you are pointing it in the exact opposite direction. The angle between $\vec{P}$ and the original $\vec{Q}$ was $\alpha$. The angle between $\vec{P}$ and the new $-\vec{Q}$ is the supplementary angle, which is $180^\circ - \alpha$. It is a beautiful geometric consequence of the straight line!

Q: Is it possible for $P$ to be negative?

In this context, $P$ represents the magnitude of a force. By definition, the magnitude of a vector is always a non-negative scalar. Even if the algebra gives you $P^2 = 25$, we take the positive root $P = 5$ because a force magnitude cannot be negative.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the two forces as vectors originating from the same point. The resultant is the diagonal of the parallelogram formed by these vectors. When you reverse the direction of the 3N force, you are essentially rotating it by 180 degrees, which changes the angle between the two vectors from

α

18 0^{\circ} - α

Hint 2

Hint 2: Utilize the Law of Cosines for vector addition, which states that the square of the resultant magnitude

R

is given by

R^{2} = A^{2} + B^{2} + 2 A B cos α

. Apply this formula to both scenarios: the initial case with angle

α

and the second case with angle

18 0^{\circ} - α

Hint 3

Hint 3: Set up your two equations:

49 = P^{2} + 9 + 6 P cos α

and

19 = P^{2} + 9 + 6 P cos (18 0^{\circ} - α)

. Recall the trigonometric identity

cos (18 0^{\circ} - α) = - cos α

. By adding these two equations, the term involving

cos α

will cancel out, allowing you to solve directly for

P^{2}

Visualized Solution

$I ni t ia l S e t u p$

Let the two forces be $P$ and $Q$ with magnitude $3$ .
Let the angle between them be $α$ .
The magnitude of their resultant is $R_{1} = 7$ .

$V ec t or A dd i t i o n F or m u l a$

Using the Law of Cosines for vector addition:
$R^{2} = A^{2} + B^{2} + 2 A B cos α$

$S u b s t i t u t in g C a se 1$

Substitute $R_{1} = 7$ , $A = P$ , and $B = 3$ :
$7^{2} = P^{2} + 3^{2} + 2 (P) (3) cos α$
$49 = P^{2} + 9 + 6 P cos α \dots (1)$

$R e v er s in g t h e 3 N F or ce$

The direction of the $3 N$ force is reversed.
The new angle between $P$ and $- Q$ becomes $18 0^{\circ} - α$ .
The new resultant is $R_{2} = 19$ .

$S u b s t i t u t in g C a se 2$

Substitute $R_{2} = 19$ , $A = P$ , and $B = 3$ :
$(19)^{2} = P^{2} + 3^{2} + 2 (P) (3) cos (18 0^{\circ} - α)$

$S im pl i f y in g E q u a t i o n 2$

Recall the trigonometric identity: $cos (18 0^{\circ} - α) = - cos α$
$19 = P^{2} + 9 - 6 P cos α \dots (2)$

$A dd in g t h e E q u a t i o n s$

Add Equation (1) and Equation (2) to eliminate $α$ :
$(49) + (19) = (P^{2} + 9 + 6 P cos α) + (P^{2} + 9 - 6 P cos α)$
$68 = 2 P^{2} + 18$

$S o l v in g f or P$

Subtract $18$ from both sides:
$2 P^{2} = 68 - 18$
$2 P^{2} = 50$
Divide by $2$ :
$P^{2} = 25$

$F ina l A n s w er$

Taking the square root:
$P = 25 = 5$
Final Answer: The magnitude of force $P$ is $5 N$ .

00:00 / 00:00

The Sigma Insight: Addition of Vectors

Solution Diagram

Analyzing the Setup

We are given two forces,

P

and

Q

, where the magnitude of

Q

3 N

. They act at an angle

α

to produce a resultant magnitude of

7 N

Using the Parallelogram Law of Forces, the resultant

R

is given by the formula:

R^{2} = P^{2} + Q^{2} + 2 P Q cos α

Substituting the known values

R = 7

and

Q = 3

, we obtain:

7^{2} = P^{2} + 3^{2} + 2 (P) (3) cos α

Simplifying this expression, we arrive at our first governing equation:

49 = P^{2} + 9 + 6 P cos α — (Equation 1)

The Twist

A Reversal of Fortune

The problem states that the force

Q

is reversed, meaning its direction changes by

18 0^{\circ}

. The new angle between

P

and the reversed force becomes

18 0^{\circ} - α

The new resultant magnitude is given as

19 N

. Applying the Law of Cosines again:

(19)^{2} = P^{2} + 3^{2} + 2 (P) (3) cos (18 0^{\circ} - α)

Using the trigonometric identity

cos (18 0^{\circ} - α) = - cos α

, the equation simplifies to:

19 = P^{2} + 9 - 6 P cos α — (Equation 2)

The Elegance of Symmetry

We now have a system of two equations: 1)

49 = P^{2} + 9 + 6 P cos α

19 = P^{2} + 9 - 6 P cos α

To isolate

P

, we add Equation (1) and Equation (2). This eliminates the term involving

cos α

(49 + 19) = (P^{2} + 9 + 6 P cos α) + (P^{2} + 9 - 6 P cos α)

This simplifies to:

68 = 2 P^{2} + 18

Final Calculation

Subtracting

18

from both sides of the equation yields:

2 P^{2} = 50

Dividing by

2

, we find:

P^{2} = 25

Taking the square root, we determine the magnitude of the unknown force: $P = 5 N$

Common Traps & FAQs

Wait, why does reversing the vector change the angle to

18 0^{\circ} - α

Imagine two vectors

P

and

Q

with an angle

α

between them. If you reverse

Q

to become

- Q

, you are pointing it in the exact opposite direction. The angle between

P

and the original

Q

was

α

. The angle between

P

and the new

- Q

is the supplementary angle, which is

18 0^{\circ} - α

. It is a beautiful geometric consequence of the straight line!

Can I use the triangle law instead of the parallelogram law?

Absolutely! The triangle law and the parallelogram law are mathematically equivalent. Whether you shift the vectors to form a triangle or keep them at a common origin to form a parallelogram, the resultant vector remains the same. The Law of Cosines

R^{2} = A^{2} + B^{2} + 2 A B cos α

is derived from both, so you will arrive at the exact same equations.

Why did we add the equations instead of subtracting them?

We added them because our goal was to eliminate the unknown variable

α

. In the first equation, we had a

+ 6 P cos α

term, and in the second, we had a

- 6 P cos α

term. Adding them makes these terms sum to zero, which is the most efficient path to isolating

P

Is it possible for

P

to be negative?

In this context,

P

represents the magnitude of a force. By definition, the magnitude of a vector is always a non-negative scalar. Even if the algebra gives you

P^{2} = 25

, we take the positive root

P = 5

because a force magnitude cannot be negative.

Animated Solution for Mathematics - Vector Algebra: The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be $19 n$ . The value of P is

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

$V ec t or A dd i t i o n F or m u l a$

$S u b s t i t u t in g C a se 1$

$R e v er s in g t h e 3 N F or ce$

$S u b s t i t u t in g C a se 2$

$S im pl i f y in g E q u a t i o n 2$

$A dd in g t h e E q u a t i o n s$

$S o l v in g f or P$

$F ina l A n s w er$

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Twist

The Elegance of Symmetry

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vector Addition Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Case 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reversing the 3N Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Case 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Equation 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adding the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Twist

The Elegance of Symmetry

Final Calculation

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$I ni t ia l S e t u p$

$V ec t or A dd i t i o n F or m u l a$

$S u b s t i t u t in g C a se 1$

$R e v er s in g t h e 3 N F or ce$

$S u b s t i t u t in g C a se 2$

$S im pl i f y in g E q u a t i o n 2$

$A dd in g t h e E q u a t i o n s$

$S o l v in g f or P$

$F ina l A n s w er$

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