If a simply-supported, static cantilever beam at equilibrium experiences a uniform downward force of 200 N/m across its entire 10m length, find the reaction force and moment at the support point of the cantilever (where the cantilever comes out of the wall).
Total force exerted on cantilever beam = -200 N/m * 10m = -2000 N (downward) Therefore, since the beam is at equilibrium, the sum of forces on the beam must be zero, so the reaction force, R, at the point of support must satisfy R - 2000 = 0 so R = 2000 N (upward) To find the reaction moment, also use the fact that the sum of the moments on the beam must be zero since the beam is at equilibrium. The moment on the beam due to the distributed load is equal to the magnitude of the total force applied multiplied by the distance of the point at which this force acts (midpoint of the beam) from the point of support, which is 5m. Moment due to load = -2000 N * 5m = -10000 Nm (clockwise) Therefore, the reaction moment, M, at the point of support must satisfy M - 10000 = 0, so M = 10000 Nm (counterclockwise)
Integrate sin(2x) - ln(x^3) from 1 to 3.
Start with the integral of sin(2x): sin(2x)dx = (1/2)*sin(u)du -(1/2)*cos(u) + C = -(1/2)*cos(2x) + C Now evaluate the integral of ln(x^3): ln(x^3)dx = 3*ln(x)dx Use integration by parts on ln(x) (f*dg = f*g - g*df), with f = ln(x), df = (1/x)dx, g = x, dg = dx: 3*ln(x)dx = 3*(x*ln(x) - dx) = 3x*(ln(x) - 1) Now evaluate -(1/2)*cos(2x) - 3x*(ln(x) - 1) from 1 to 3: -(1/2)*(cos(6) - cos(2)) - (9*(ln(3) - 1) - 3*(ln(1) - 1)) = (after much simplification) = sin(2)*sin(4) + 6 - 9*ln(3), which approximately equals -4.58
From rest, a ball is dropped straight down from a height of 10m on Earth. How long does it take the ball to hit the ground?
v_0 = 0 m/s a = 9.8 m/s^2 d = 10 m d = v_0*t + (1/2)*a*t^2 10 = (1/2)*9.8*t^2 t^2 = 10/(4.9) t^2 = 2.041 t = 1.43 seconds