Two forces, F1 and F2 act on an object and the angle between them is 60 degrees. Determine the magnitude and the direction of the resultant force FR relative to F1 if F1= 50 N and F2= 30 N.
Did you make a diagram?
I drew the 20 N line, and at its end drew the 50 N line with an angle of 120° between them.
(you should be familiar with the parallelogram resulting from the addition of 2 vectors with the diagonal representing the resultant R)
|R|^2 = 50^2 + 20^2 - 2(50)(20)cos 120°
= 2500 + 400 - 2000(-1/2)
= 3900
|R| = √3900 = appr. 62.45 N
now you can use the Sine Law to find the angle between either vector and the resultant.
To determine the magnitude and direction of the resultant force FR, we can use the law of cosines, which states:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, F1 is force "a" with a magnitude of 50 N, F2 is force "b" with a magnitude of 30 N, and the angle between them is 60 degrees (C).
Now, let's substitute the values into the equation:
FR^2 = F1^2 + F2^2 - 2 * F1 * F2 * cos(60)
FR^2 = (50^2) + (30^2) - 2 * 50 * 30 * cos(60)
FR^2 = 2500 + 900 - 3000 * 0.5
FR^2 = 2500 + 900 - 1500
FR^2 = 2900 - 1500
FR^2 = 1400
Taking the square root of both sides:
FR ≈ √1400
FR ≈ 37.42 N (rounded to two decimal places)
So, the magnitude of the resultant force FR is approximately 37.42 N.
To find the direction of FR relative to F1, we can use the law of sines:
sin(A) / a = sin(B) / b = sin(C) / c
In this case, A is the angle between FR and F1, which we want to find.
Let's use sin(A) / F1 = sin(60) / FR:
sin(A) / 50 = sin(60) / 37.42
Cross multiply:
sin(A) = 50 * sin(60) / 37.42
sin(A) ≈ 0.866
Using the inverse sine function, we find:
A ≈ 60.65 degrees
Therefore, the direction of the resultant force FR relative to F1 is approximately 60.65 degrees.
To determine the magnitude and direction of the resultant force, we can use the law of cosines. The law of cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the lengths of those two sides multiplied by the cosine of the included angle.
In this case, we have a triangle with sides of lengths F1, F2, and FR, and an included angle of 60 degrees between F1 and F2. We can label the angle opposite side FR as angle A.
Using the law of cosines, we can write:
FR^2 = F1^2 + F2^2 - 2 * F1 * F2 * cos(A)
Plugging in the given values:
FR^2 = 50^2 + 30^2 - 2 * 50 * 30 * cos(60)
Simplifying:
FR^2 = 2500 + 900 - 3000 * 0.5
FR^2 = 2500 + 900 - 1500
FR^2 = 2900
Taking the square root of both sides, we find:
FR ≈ 53.85 N
So the magnitude of the resultant force, FR, is approximately 53.85 N.
To determine the direction of the resultant force, we can use the law of sines. The law of sines states that in a triangle, the ratio of the length of a side to the sine of the angle opposite that side is the same for all sides and angles in the triangle.
In this case, we can label the angle opposite side F1 as angle B.
Using the law of sines, we can write:
sin(A) / F1 = sin(B) / FR
Plugging in the given values:
sin(60) / 50 = sin(B) / 53.85
Simplifying:
0.866 / 50 = sin(B) / 53.85
Multiplying both sides by 53.85:
sin(B) ≈ 0.934
Taking the inverse sine of 0.934, we find:
B ≈ 68.57 degrees
Therefore, the direction of the resultant force, FR, relative to F1 is approximately 68.57 degrees.