find by calculation,the resultant two equal forces of 6N each acting at a 4N point at an angle of (a) 60 degree (b)120 degree with each other{hint cos120= -cos60 =1 and half or 1/2
cos 60 =V2
what is a "4N point" ?
and your cosine hints are mangled.
To find the resultant of two equal forces acting at an angle, we can use the parallelogram law of vector addition.
(a) For an angle of 60 degrees:
Step 1: Draw a diagram representing the two forces. We have two equal forces of 6N each acting at a 4N point, and the angle between them is 60 degrees.
Step 2: Draw a parallelogram using the two vectors. Place the first vector at the origin and the second vector at the endpoint of the first vector.
Step 3: Complete the parallelogram by drawing lines parallel to the vectors.
Step 4: Measure the length of the diagonal of the parallelogram. This diagonal represents the resultant force.
Step 5: Use the Pythagorean theorem to find the magnitude of the resultant force:
Resultant force (R) = √(6N)^2 + (6N)^2 + 2(6N)(6N)cos(60°)
R = √(36N^2 + 36N^2 + 72N^2(1/2))
R = √(72N^2 + 72N^2 + 72N^2)
R = √(216N^2)
R = 6√6N
The magnitude of the resultant force for an angle of 60 degrees is 6√6N.
(b) For an angle of 120 degrees:
Step 1: Follow steps 1-3 from part (a).
Step 2: The diagonals of a parallelogram bisect each other. Draw a line from the intersection of the diagonals to the origin. This line represents the resultant force.
Step 3: Measure the length of this line to find the magnitude of the resultant force.
Step 4: Use the Law of Cosines to find the magnitude of the resultant force:
Resultant force (R) = √(6N)^2 + (6N)^2 - 2(6N)(6N)cos(120°)
R = √(36N^2 + 36N^2 - 72N^2(-1/2))
R = √(72N^2 + 72N^2 + 72N^2)
R = √(216N^2)
R = 6√6N
The magnitude of the resultant force for an angle of 120 degrees is 6√6N.
In both cases, the resultant force is 6√6N.