Find the magnitude and direction of
a)the resultant
b)the equilibrant of two forces 10N acting in the direction N030'E and 15N acting in the eastern direction if boths forces act at a point.
Please solve nd give the details especially the B parts of the question.
We are not going to do your work for you. Do you have a question on it? You add the forces as vectors.
The equilibrant is equal and in opposite direction to the resulatant.
equilibrant= - resultant
jlj
To find the magnitude and direction of the resultant and equilibrant of two forces, we need to use vector addition and subtraction.
a) Resultant of two forces:
To find the resultant, we need to add the two forces together. The magnitude of the resultant is the sum of the magnitudes of the two forces, and the direction is the angle at which the resultant points.
Given:
Force 1: 10N, direction N030'E
Force 2: 15N, direction east
Step 1: Convert the given directions into a common reference frame.
N030'E can be converted to an angle with respect to the positive x-axis:
N030'E = 30 degrees clockwise from the north direction (or counter-clockwise from the east direction).
Step 2: Convert the forces into their vector component form.
Force 1: 10N at 30 degrees from the positive x-axis:
Fx1 = 10N * cos(30) = 8.660N (rounded to three decimal places)
Fy1 = 10N * sin(30) = 5.000N
Force 2: 15N in the positive x-direction:
Fx2 = 15N
Fy2 = 0N
Step 3: Add the x and y components of the forces separately.
Resultant x-component: Fx = Fx1 + Fx2 = 8.660N + 15N ≈ 23.660N
Resultant y-component: Fy = Fy1 + Fy2 = 5.000N + 0N = 5.000N
Step 4: Find the magnitude and direction of the resultant.
Magnitude of the resultant: |R| = sqrt(Fx^2 + Fy^2) = sqrt((23.660N)^2 + (5.000N)^2) ≈ 24.220N
Direction of the resultant: θ = arctan(Fy / Fx) = arctan(5.000N / 23.660N) ≈ 12.074 degrees
Therefore, the magnitude of the resultant force is approximately 24.220N, and its direction is approximately 12.074 degrees.
b) Equilibrant of two forces:
To find the equilibrant force, we need to subtract one force from the other. The magnitude of the equilibrant is equal to the magnitude of the resultant, but its direction is the opposite of the resultant force.
Given:
Force 1: 10N, direction N030'E
Force 2: 15N, direction east
Step 1: Convert the given directions into a common reference frame (same as in part a).
Step 2: Convert the forces into their vector component form (same as in part a).
Step 3: Subtract the x and y components of one force from the other.
Equilibrant x-component: Ex = Fx2 - Fx1 = 15N - 8.660N ≈ 6.340N
Equilibrant y-component: Ey = Fy2 - Fy1 = 0N - 5.000N = -5.000N
Step 4: Find the magnitude and direction of the equilibrant.
Magnitude of the equilibrant: |E| = sqrt(Ex^2 + Ey^2) = sqrt((6.340N)^2 + (-5.000N)^2) ≈ 8.028N
Direction of the equilibrant: θ = arctan(-Ey / Ex) = arctan(5.000N / 6.340N) ≈ 40.995 degrees
Therefore, the magnitude of the equilibrant force is approximately 8.028N, and its direction is approximately 40.995 degrees.
not that i can't solve,i hav already solve 4 d resultant to be 21.8N.after using cosine rule..
But am confused in solving for the equilibrant dat is the B parts of d question Mr bobpursley,how can i get the equilibrant.