Lesson 1, Topic 3
In Progress

# Parallelogram Law of Vectors (Alternate Method)

Lesson Progress
0% Complete

Recall,

If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.

Figure 1 represents two vectors acting at a point O at an angle of θ with each other.

Complete the parallelogram by drawing DE parallel to OB and OD parallel to BE as shown in figure 2 and join OE. OE represents the resultant vector R in magnitude and direction.

Let OB represent vector B and OD represent vector D.

∠O and ∠D are suplementary

∴ ∠O + ∠D = 180°

θ + ∠D = 180°

∠D = 180 – θ

Considering figure 3, the resultant can be found using cosine rule;

R2 = (OB)2 + (OD)2 + 2(OB)(OD) Cos(180 – θ)°

R2 = B2 + D2 + 2BD Cos(180 – θ)°

where θ is the range between the vectors.

### Direction of the Resultant Vector:

The direction of a resultant vector is the angle which the resultant vector makes with the horizontal at the point of its origin $$\left(\scriptsize \alpha \right)$$.

The direction of resultant vector can be calculated using;

$$\frac{B}{sinB} = \frac{D}{sinD} = \frac{R}{sinR}$$

The formula chosen depends on the angles whose sides and angles are given.

Let’s take a look at some examples;

### Example 1:

Forces of 22 N and 26 N act on the same object at an angle of 47° to each other. Find the magnitude and direction of the resultant vector.

Solution:

We can sketch the diagram as shown below.

To find the resultant redraw the diagram as shown below.

Use the cosine rule to figure out the magnitude of R.

R2 = (22)2 + (26)2 – 2(22)(26) Cos(180 – 47)°

R2 = (22)2 + (26)2 – 2(22)(26) Cos(133)°

R2 = 484 + 676 – 1,144 x (-0.682)

R2 = 1160 + 780.2

R2 = 1940.2

R = $$\scriptsize \sqrt{1940.2}$$

R = 44.048N

magnitude of the resultant vector = 44.048N

To find the direction of the resultant $$\left(\scriptsize \alpha \right)$$ we will use the sine rule

$$\frac{sin \: \alpha}{22} = \frac{sin \: 133}{44.048}$$

cross multiply

$$\scriptsize sin \: \alpha \: \times \: 44.048 = sin \: 133^o\: \times \: 22$$

$$\scriptsize sin \: \alpha = \normalsize \frac{sin \: 133^o\: \times \: 22}{44.048}$$

$$\scriptsize sin \: \alpha = \normalsize\frac{0.7314\: \times \: 22}{44.048}$$

$$\scriptsize sin \: \alpha = \normalsize \frac{16.0908}{44.048}$$

$$\scriptsize sin \: \alpha = 0.3653$$

$$\scriptsize \alpha = sin^{-1}(0.3653)$$

$$\scriptsize \alpha = 21.4^o$$

Direction of the vector = 21.4°

### Example 2:

Two forces with magnitude of 15 N and 35 N are applied to an object as shown in the diagram below. The magnitude of the resultant is 28 N. Find the measurement of the angle between the resultant vector and the vector of the 15 N force to the nearest whole degree.

Solution:

Complete the parallelogram by drawing lines parallel to the two forces and draw the resultant vector as shown below. x is the angle between the resultant vector and the vector of the 15 N force.

Using the cosine rule:

352 = 152 + 282 – 2(15)(28)cosx

1225 = 225 + 784 – 840cosx

-840cosx = 1225 – 1009

cosx = $$\frac{216}{-840}$$

cosx = -0.257

x = cos-1(-0.257)

x = 104.9°

#### Responses error: