Resolving vectors into components

[Gringo123]

I have just looked at a problem in an A level physics book which asks you to work out the forward force acting on a boat. There are 7 N of force going straight ahead and a further 5 N of force going 30 degrees to the left. In order to work out the forward force of the boat the book says this:

First you need to find the amount of the 5 N force that acts in the forward direction, using trigonometry:

Part of 5 N force in forward direction = 5 cos 30° = 4.3 N

Then this can be added to the 7 N force:

4.3 + 7 = 11.3 N force in the forward direction.

How do I work out that 5 cos 30° = 4.3 N?

 

[phyzmatix]

It stems from the trig identity

$$\cos \theta = \frac{x}{r}$$

In this case, x represents your forward direction (draw yourself a sketch to see) and r represents the force vector. The rest is then simply algebraic manipulation of this identity and a final application of the trigonometric rules to find the value of x.

Makes sense?

 

[sir-pinski]

To understand how to work out the amount of the 5 N force in the forward direction, we need to first understand how vectors can be resolved into components. In this problem, we have two forces acting on the boat - 7 N going straight ahead and 5 N going 30 degrees to the left. These two forces can be represented by vectors, with the direction and magnitude of the force indicated by the length and direction of the arrow.

To resolve these vectors into components, we need to break them down into their horizontal and vertical components. The horizontal component is the part of the force that acts in the horizontal direction, while the vertical component is the part of the force that acts in the vertical direction. In this case, we are interested in finding the horizontal component of the 5 N force, as that will give us the amount of force in the forward direction.

To find the horizontal component, we can use trigonometry. In a right-angled triangle, the cosine of an angle is equal to the adjacent side divided by the hypotenuse. In this case, the adjacent side is the horizontal component of the 5 N force, and the hypotenuse is the 5 N force itself. So, by using the formula cos 30° = adjacent/hypotenuse, we can solve for the adjacent side, which represents the horizontal component of the 5 N force.

Substituting the values in the formula, we get:

cos 30° = adjacent/5

Rearranging the equation, we get:

adjacent = 5 cos 30°

Plugging in the value of cos 30° (which is 0.866) into the equation, we get:

adjacent = 5 x 0.866 = 4.3 N

Therefore, the amount of the 5 N force in the forward direction is 4.3 N. Adding this to the 7 N force gives us a total of 11.3 N force in the forward direction.

In summary, to work out the amount of the 5 N force in the forward direction, we use trigonometry to find the horizontal component of the force. This is done by using the formula cos 30° = adjacent/hypotenuse and solving for the adjacent side, which represents the horizontal component. By plugging in the values and solving the equation, we get the amount of the 5 N force in the forward direction as