A particle weighing 69 newtons is placed on a plane inclined at an angle 𝜃 to the horizontal, where the tan of 𝜃 is equal to four-thirds. Resolve the weight of the particle into two vector components, 𝐅 sub one and 𝐅 sub two, where 𝐅 sub one is parallel to a line of greatest slope and 𝐅 sub two is perpendicular to 𝐅 sub one. And find their intensities 𝐅 sub one and 𝐅 sub two.
We will begin by sketching a diagram showing the particle placed on an inclined plane. We are told that the particle weighs 69 newtons. This means that there will be a force acting vertically downwards equal to 69 newtons. The plane is inclined at an angle of 𝜃 to the horizontal. And we are told that the tan of angle 𝜃 is equal to four-thirds. We will return to this later.
We are asked to resolve the weight of the particle into two vector components 𝐅 sub one and 𝐅 sub two as shown on the diagram. 𝐅 sub one is parallel to the line of greatest slope, and 𝐅 sub two is perpendicular to 𝐅 sub one. We can find the intensities or magnitudes of these components using our knowledge of right angle trigonometry. 𝐅 sub one is equal to the magnitude of the weight force 69 newtons multiplied by the sin of angle 𝜃. And 𝐅 sub two is equal to 69 multiplied by cos 𝜃.
Since the tan of any angle 𝜃 is equal to the opposite over the adjacent and we know this is equal to four-thirds, we can use our knowledge of the Pythagorean theorem or Pythagorean triples to calculate the value of sin 𝜃 and cos 𝜃. Since the hypotenuse of the triangle drawn is equal to five units, the sin of angle 𝜃 is equal to four-fifths and the cos of angle 𝜃 three-fifths.
𝐅 sub one is therefore equal to 69 multiplied by four-fifths, and this is equal to 55.2 newtons. 𝐅 sub two is equal to 69 multiplied by three-fifths, and this is equal to 41.4 newtons. The intensities of 𝐅 sub one and 𝐅 sub two are 55.2 and 41.4 newtons, respectively.