(a) Find the resultant of the following coplanar forces acting at point O: 44N in direction 54°, 77N in direction 90°, 60N in direction 146°, 88N in direction 245°. Directions are measured anticlockwise from the positive x-axis.
(b) If y = (x2 – 2x + 5)(3x +2x3) find dy/dx. Do not simplify.
a. F = 44N[54o] + 77[90] + 60[146] + 88[245].
X = 44*Cos54+77*Cos90+60*Cos146+88*Cos245 = Sum of hor. components.
Y = 44*sin54+77*sin90+60*sin146+88*sin245 = Sum of ver. components.
Tan A = Y/X, A = ?
Resultant = X/Cos A or Sqrt(X^2+Y^2).
(a) To find the resultant of the coplanar forces acting at point O, we need to determine the sum of the forces in both the x and y directions.
1. Resolve each force into its x and y components:
- Force of 44N at 54°: F1x = 44N * cos(54°), F1y = 44N * sin(54°)
- Force of 77N at 90°: F2x = 77N * cos(90°), F2y = 77N * sin(90°)
- Force of 60N at 146°: F3x = 60N * cos(146°), F3y = 60N * sin(146°)
- Force of 88N at 245°: F4x = 88N * cos(245°), F4y = 88N * sin(245°)
2. Calculate the sum of the x and y components:
Rx = F1x + F2x + F3x + F4x
Ry = F1y + F2y + F3y + F4y
3. Find the magnitude and direction of the resultant:
Resultant force, R = sqrt(Rx^2 + Ry^2)
Direction, θ = arctan(Ry / Rx)
Hence, the resultant of the coplanar forces acting at point O is R N in the direction of θ°.
(b) To find dy/dx for the given function y = (x^2 – 2x + 5)(3x + 2x^3), we will differentiate it using the product rule of differentiation.
1. Apply the product rule:
dy/dx = (3x + 2x^3) * d/dx(x^2 – 2x + 5) + (x^2 – 2x + 5) * d/dx(3x + 2x^3)
2. Differentiate the individual terms:
d/dx(x^2 – 2x + 5) = 2x - 2
d/dx(3x + 2x^3) = 3 + 6x^2
3. Substitute the differentiated terms into the product rule equation:
dy/dx = (3x + 2x^3)(2x - 2) + (x^2 – 2x + 5)(3 + 6x^2)
Hence, the expression for dy/dx is the resulting equation after applying the product rule to the given function y.