This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Morning Lïský — let's get this done. To solve this problem, we will apply the conditions for static equilibrium: the sum of forces in the x-direction is zero, the sum of forces in the y-direction is zero, and the sum of torques about any point is zero. Given: Length of ladder (L) = 10 m Weight of ladder (W_L) = 200 N (acts at L/2 = 5 m from the base) Angle with horizontal () = 60^ Weight of boy (W_B) = 500 N (acts at L/4 = 2.5 m from the base) The wall is smooth, so there is no friction at the wall. Let N_W be the normal reaction at the wall, N_G be the normal reaction from the ground, and F_f be the friction force from the ground. Step 1: Calculate the normal reaction at the wall (N_W). We take moments (torques) about the lower end of the ladder (point B in the diagram). The forces N_G and F_f pass through this point, so they create no torque. The clockwise torques are due to the weights of the ladder and the boy. The anti-clockwise torque is due to the normal reaction from the wall. Horizontal distance for W_L from B = 5 60^. Horizontal distance for W_B from B = 2.5 60^. Vertical distance for N_W from B = 10 60^. For equilibrium, the sum of torques is zero: W_L (5 60^) + W_B (2.5 60^) = N_W (10 60^) Substitute the given values: 200 N (5 × 0.5) + 500 N (2.5 × 0.5) = N_W (10 × sqrt(3)2) 200 N (2.5) + 500 N (1.25) = N_W (5sqrt(3)) 500 Nm + 625 Nm = N_W (5sqrt(3) m) 1125 Nm = N_W (5sqrt(3) m) N_W = (1125)/(5sqrt(3)) N N_W = (225)/(sqrt(3)) N Rationalize the denominator: N_W = 225sqrt(3)3 N N_W = 75sqrt(3) N The normal reaction at the wall is 75sqrt(3) N. Step 2: Calculate the forces at the lower end of the ladder. These forces are the normal reaction from the ground (N_G) and the friction force from the ground (F_f). Sum of vertical forces ( F_y = 0): N_G - W_L - W_B = 0 N_G = W_L + W_B N_G = 200 N + 500 N N_G = 700 N Sum of horizontal forces ( F_x = 0): Since the wall is smooth, there is no friction at the wall. The only horizontal forces are the friction from the ground (F_f) and the normal reaction from the wall (N_W). F_f - N_W = 0 F_f = N_W F_f = 75sqrt(3) N Step 3: Calculate the magnitude of the resultant force acting on the lower end. The resultant force (F_R) at the lower end is the vector