Two forces of 5kg wt and 10kg wt are acting at an inclination of 120
o
between them.
Find the magnitude and direction of resultant force.
Since you don't specify either force's direction,
let the 5N force act in the x direction. Its vector value is then (5,0)
The other force is then (-5,5√3)
Add them together to get (0,5√3)
That's 5√3 at right angle to the N force.
To find the magnitude and direction of the resultant force, we need to use the parallelogram law of forces. According to this law, if two forces are acting at an inclination to each other, the resultant force can be found by constructing a parallelogram using the two forces as adjacent sides. The diagonal of the parallelogram represents the magnitude and direction of the resultant force.
First, we need to find the vector components of the given forces. Let's assume that the 5kg wt force is F1 and the 10kg wt force is F2.
To find the x-component of F1, we use the equation:
Fx1 = F1 * cos(120°)
To find the y-component of F1, we use the equation:
Fy1 = F1 * sin(120°)
Since the magnitude of F1 is given as 5kg wt, we can substitute that in the equations:
Fx1 = 5kg wt * cos(120°)
Fy1 = 5kg wt * sin(120°)
Calculating these values:
Fx1 = -2.5kg wt
Fy1 = 4.330kg wt
Now, similarly, let's find the vector components of F2:
Fx2 = F2 * cos(120°)
Fy2 = F2 * sin(120°)
Substituting the given magnitude of F2 (10kg wt) in the equations:
Fx2 = 10kg wt * cos(120°)
Fy2 = 10kg wt * sin(120°)
Calculating these values:
Fx2 = -5kg wt
Fy2 = 8.660kg wt
Now that we have the components of both forces, we can add them to find the resultant components:
Rx = Fx1 + Fx2
Ry = Fy1 + Fy2
Substituting the values we calculated:
Rx = -2.5kg wt + (-5kg wt) = -7.5kg wt
Ry = 4.330kg wt + 8.660kg wt = 12.99kg wt (approximately)
The magnitude of the resultant force, R, can be calculated using the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
Substituting the values:
R = sqrt((-7.5kg wt)^2 + (12.99kg wt)^2) = sqrt(56.25kg wt + 168.76kg wt) = sqrt(225kg wt + 168.76kg wt) = sqrt(393.76kg wt) = 19.84kg wt (approximately)
To find the direction of the resultant force, we can use the tangent of the angle theta, which is given by:
theta = atan(Ry / Rx)
Substituting the values:
theta = atan(12.99kg wt / -7.5kg wt) = atan(-1.732) = -60° (approximately)
Therefore, the magnitude of the resultant force is approximately 19.84kg wt, and its direction is -60° (or 300°) relative to the positive x-axis.
To find the magnitude and direction of the resultant force, we can use the concept of vector addition.
First, let's convert the weights (kg wt) into Newtons (N) since force is typically measured in Newtons. We know that 1 kg wt is equal to 9.8 N, so we can calculate the forces as follows:
Force1 = 5 kg wt * 9.8 N/kg wt = 49 N
Force2 = 10 kg wt * 9.8 N/kg wt = 98 N
Next, we can represent these forces as vectors. We'll use a coordinate system where the positive x-axis is in the direction of Force1, and the positive y-axis is perpendicular to the x-axis in the plane of the forces.
Let's denote Force1 as vector F1 and Force2 as vector F2. We can express these vectors in terms of their x and y components:
F1x = Force1 * cos(120°)
F1y = Force1 * sin(120°)
F2x = Force2
F2y = 0 (since Force2 acts along the x-axis)
Using the given inclination angle of 120°, we can deduce that:
F1x = -49 * cos(120°)
F1y = 49 * sin(120°)
Now, we can find the resultant force vector by adding the individual components:
Resultant force, F = F1 + F2
Fresultantx = F1x + F2x
Fresultanty = F1y + F2y
Fresultantx = -49 * cos(120°) + 98
Fresultanty = 49 * sin(120°)
To find the magnitude (R) and direction (θ) of the resultant force, we can use the Pythagorean theorem and trigonometry:
R = sqrt( Fresultantx^2 + Fresultanty^2 )
θ = tan^(-1) ( Fresultanty / Fresultantx )
By substituting the calculated values, we can find the magnitude and direction of the resultant force.