Two forces act at an angle of 120°,the bigger force is of 80newton and the resultant is perpendicular to the smaller one.find the smaller one.
Let's assume that the smaller force is represented by vector A and the larger force is represented by vector B. We are given that the magnitude of vector B is 80 Newtons and that the angle between the two forces is 120°.
We can use vector addition to find the resultant of the two forces. If the resultant is perpendicular to vector A, it means that the two forces are parallel and in opposite directions.
First, we represent vector A using Cartesian coordinates: A = (Ax, Ay). Since the resultant is perpendicular to vector A, its x-component must be equal in magnitude but opposite in direction to the x-component of vector A. Therefore, the x-component of the resultant is -Ax.
Next, we represent vector B using Cartesian coordinates: B = (80cos(120°), 80sin(120°)).
Now, we can find the x-component of the resultant by adding the x-components of vector A and vector B: -Ax = 80cos(120°) + Ax
Simplifying this equation: 2Ax = 80cos(120°)
Ax = 40cos(120°) = -20.
Since vector A is represented by (Ax, Ay), we know that Ay must equal zero for the resultant to be perpendicular to vector A.
Thus, the smaller force represented by vector A is 20 Newtons.
To find the magnitude of the smaller force, we can use the concept of vector addition and trigonometry.
Step 1: Using the given information, we know that the angle between the two forces is 120° and the magnitude of the bigger force is 80 Newtons.
Step 2: Let's assume the magnitude of the smaller force as F (unknown).
Step 3: Since the resultant force is perpendicular to the smaller force, we can conclude that the two forces form a right-angled triangle.
Step 4: Using trigonometry, we can determine the relationship between the magnitudes of the forces and the angle between them in a right-angled triangle. In this case, we will use the trigonometric ratio cosine (cos).
cos(120°) = Adjacent side / Hypotenuse
In this case, the adjacent side is the magnitude of the bigger force (80 Newtons), and the hypotenuse will be the magnitude of the resultant.
cos(120°) = 80 / Resultant force (R)
Step 5: Since the cosine of 120° is equal to -0.5, we can solve for the magnitude of the resultant force:
-0.5 = 80 / R
Step 6: Cross multiplying the equation, we get:
-0.5R = 80
Step 7: Dividing both sides of the equation by -0.5, we find:
R = 80 / -0.5
≈ -160 Newtons
Step 8: However, force can only be positive, so we take the absolute value of the magnitude of the resultant:
|R| ≈ |-160|
≈ 160 Newtons
Therefore, the magnitude of the smaller force is approximately 160 Newtons.
To find the magnitude of the smaller force, we can use vector addition and trigonometry.
First, let's represent the two forces as vectors. Let F1 be the bigger force of 80 Newtons, and F2 be the smaller force we are trying to find.
Next, we can use the concept of vector addition to find the resultant force. The resultant force is the vector sum of F1 and F2. Since the resultant force is perpendicular to the smaller force (F2), we know that the angle between F1 and F2 is 90 degrees.
Now, let's break down the forces into their horizontal and vertical components. The horizontal component of F1 is equal to F1 times the cosine of the angle between F1 and the resultant force. Since the angle between F1 and the resultant is 120 degrees, the angle between F2 and the resultant force is 120 - 90 = 30 degrees. Therefore, the horizontal component of F1 is:
Horizontal component of F1 = F1 * cos(120°) = 80 N * cos(120°)
Similarly, the horizontal component of F2 is:
Horizontal component of F2 = F2 * cos(30°)
Since the resultant force is perpendicular to F2, the horizontal components of F1 and F2 must cancel each other out. Therefore, we have the equation:
Horizontal component of F1 + Horizontal component of F2 = 0
Simplifying this equation gives us:
80 N * cos(120°) + F2 * cos(30°) = 0
Now we can solve for F2:
F2 * cos(30°) = -80 N * cos(120°)
F2 = (-80 N * cos(120°)) / cos(30°)
Calculating this equation will give us the magnitude of the smaller force, F2.