Vector Components and Trigonometric Functions in Physics, Exercises of Physics

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• General Physics
• Module 3: Quarter 1 – Week

Jumpstart

Right triangle is one basic concept that is use in dealing with Component method. Trigonometric functions are required to solve unknown values in a right triangle. Here, we will be using the three basic trigonometric functions, namely, sine, cosine, and tangent functions. A right triangle is a triangle in which one angle is a 90^0 angle called right angle. The relation between the sides and angles of a right triangle is an important basis for trigonometry. Below is a right triangle and its parts needed in our discussion. B Where: A,B,C are the angles, angle C is 900 The sum of A and B is 900 a (^) a,b,c are the sides of the right triangle side c is called hypotenuse, which is longer than any of the two sides C A b The right triangle is our main concern in dealing with the Component method. The functions sin, cos, and tan are three functions we will be using as a pre-requisite of the component method. Using a scientific calculator, you can determine the value of an angle for a particular function used. Say, sin 25^0 , to find the value, simply press 25 and presssin in your calculator. To do the reverse, i.e., getting the angle, just press the number,then press inv/shift/2ndF, then press the function (sin-^1 , cos-^1 , tan-^1 )

Activity 1. Finding the numerical value of an angle.

Direction: Determine the value of the given angles of a certain function. Write your

answer in the blank. (Use scientific calculator in this activity)

1. cos 500 =
2. sin 200 =
3. tan 450 =
4. cos 900 =

5. sin 300 =

Activity 2. Finding the angle.

Direction: Determine the angle with their corresponding function of the

measurements. Write your answer in the blank.

1. 0.573577 sin-^1 =
2. 5.671280 tan-^1 =
3. 0.642788 cos-^1 =
4. 0.707107 sin-^1 =
5. 0.500000 cos-^1 =

Discover

The definition and equations of the three basic trigonometric functions are given below. sin θ = opposite side B hypotenuse cos θ = adjacent side a hypotenuse tan θ = opposite side C A adjacent side b Relative to angle A, <A Relative to angle A, <A sin<A = a c cos<A = b c tan<A = a b sin<B = b c cos<B = a c tan<B = b a We can also apply Pythagorean theorem in finding the value of any sides given the two sides. The equation is c^2 = a^2 + b^2 a^2 = c^2 - b^2 b^2 = c^2 - a^2

ax - horizontal component ay -^ vertical^ component

Explore

The method of component reduces all vector addition to the addition of perpendicular vectors using the Trigonometric functions. You can determine the value of x and y components of each vector, following the equations below: For component of vector (^) d x – component: y – component: dx =^ d^ cos^ θ dx =^ d^ sin^ θ Example: Given t h e f o l l o w i n g vectors, determine the value of their x and y components. d 1 = 600 m 500 N of W (^) d 2 = 400 m 750 S of E Solution: d1x = d cos (^500) d2x = d cos 750 = (600 m) (0.643) = 424.2 m = (400 m) (0.259) = 103.6 m d1y = d^ sin (^500) d2y = (^) d sin 750 = (600 m) (0.766) = 459.6 m = (400 m) (0.966) = 386.4 m

Activity 1. Finding the magnitude of the x and y components

Direction: First draw the vector diagram and resolve the components of each given

vectors. Then determine the values of the x and y components following the rules explained above. Use a separate sheet of paper for your answer

Step 2. Determine the value of the x and y components of each vector. Component of the 1 st^ vector F1x = F 1 cos 500 = (25 N) (0.643) = - 16.1 N F1y = F 1 sin 500 = (25 N) (0.766) = - 38.3 N Component of the 2 nd^ vector F2x = F 2 cos 150 = (40 N) (0.966) = 38.6 N F2y = F 2 sin 150 = (25 N) (0.259) = 10.4 N Step 3. Determine the summation or add up all components along the x-axis and y-axis. (Algebraic sum) ΣFx = F1x + F2x = (-16.1 N) + (38.6 N) = 22.5 N ΣFy = F1y + F2y = (-38.3 N) + (10.4 N) = - 27.9 N Step 4. Draw the computed sum of the x and y components in the Cartesian coordinate, using the head-to-tail method. Then draw the resultant, starting from the origin to the head of the vector along y-axis. ΣFx = 22.5 N θ FR ΣFy = 27.9 N

Step 5. Apply Pythagorean theorem to solve the magnitude of the resultant. FR = = = 35.8 N Step 6. Determine the direction angle of the resultant using tangent function. (Refer to the diagram in step 4) tan θ = ΣFy ΣFx 27.9 N tan θ = 22.5 N θ = 1.24 tan-^1 θ = 51.1^0 So the resultant of the two vector is:

FR = 35.8 N 51.1^0 S of E

Now it’s your turn!

Direction: Determine the magnitude and direction of the resultant of the three

displacements: Use separate sheet of paper for your solution. d 1 =^150 m^400 N^ of^ W d 2 =^100 m^600 N^ of^ E d 3 =^175 m^300 S^ of^ E

dR =?

5. Show the diagram of the second displacement of the car, d 2 d 2 d 2 d 2 A B^ C^ D
6. Calculate^ the^ value^ of^ the^ x - component^ of^ d 1 A. 17.3 km B. - 17.3 km C. 10 km D. - 10 km
7. Determine^ the^ value^ of^ the^ y - component^ of^ d 1 A. 17.3 km B. - 17.3 km C. 10 km D. - 10 km
8. Calculate the value of the^ x - component of^ d 2 A. - 21.2 km B. 21.2 km C. 10 km D. - 10 km
9. Determine the value of the y - component of (^) d 2 A. - 21.2 km B. 21.2 km C. 10 km D. - 10 km
10. Compute the summation of x - component A. 3.9 km B. - 3.9 km C. 31.1 km D. - 31.1 km
11. Solve the summation of y- component A. 3.9 km B. - 3.9 km C. 31.1 km D. - 31.1 km

12. Show the vector diagram of the summation of x- components, summation of y- components, (^) Σdy Σdy Σdx and A (^) B C^ D
13. Using the Pythagorean theorem, calculate the resultant displacement of the two displacements. A. 35.24 km B. - 35.24 km C. 31.34 km D. - 31.34 km
14. Determine the direction angle of the resultant displacement, (^) dR function. A. 770 B. 800 C. 830 D. 860
15. Describe the complete resultant displacement of the car. A. 35.24 km 770 S of E B. - 35.24 km 800 S of E C. 31.34 km 830 S of W D. - 31.34 km 86^0 S of E using tangent