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Okay Nike🙃, let's solve question 2. Here's the information given for the non-uniform ladder: Length of ladder L = 8.0 m Mass of ladder m = 10.0 kg Center of gravity (CG) from the bottom B: d_CG = 2.0 m Angle with the vertical wall: = 30^. This means the angle with the horizontal ground is = 90^ - 30^ = 60^. Upper end (A) against a smooth* wall (no friction). Lower end (B) on rough* ground (friction present). Acceleration due to gravity g = 9.81 m/s^2. a) Draw a free body diagram of the ladder. Imagine the ladder AB leaning against a vertical wall with its base on horizontal ground. Draw the ladder inclined at 60^ to the horizontal. Label the top end A and the bottom end B. Weight (W): A vertical arrow pointing downwards from the center of gravity G. G is located 2.0 m from B along the ladder. The magnitude is W = mg. Normal force at the wall (N_A): A horizontal arrow pointing away from the wall (to the right), acting at point A. Since the wall is smooth, there is no friction here. Normal force at the ground (N_B): A vertical arrow pointing upwards, acting at point B. Frictional force at the ground (F_f): A horizontal arrow pointing towards the wall (to the left), acting at point B. This force opposes the tendency of the ladder to slip away from the wall. b) Calculate: First, let's calculate the weight of the ladder: W = mg = (10.0 kg)(9.81 m/s^2) = 98.1 N For the ladder to be in equilibrium, the net force and net torque must be zero. Sum of forces in the horizontal (x) direction = 0: N_A - F_f = 0 N_A = F_f (Equation 1) Sum of forces in the vertical (y) direction = 0: N_B - W = 0 N_B = W = 98.1 N (Equation 2) Sum of moments about point B (bottom of the ladder) = 0: Taking moments about B eliminates N_B and F_f from the moment equation. The weight W creates a clockwise moment. The perpendicular distance from B to the line of action of W is d_CG () = (2.0 m) (60^). The normal force N_A creates an anti-clockwise moment. The perpendicular distance from B to the line of action of N_A is L () = (8.0 m) (60^). W (d_CG (60^)) = N_A (L (60^)) (98.1 N) (2.0 m × 0.5) = N_A (8.0 m × sqrt(3)2) 98.1 = N_A (4.0 sqrt(3)) N_A = (98.1)/(4.0 sqrt(3)) = (98.1)/(6.9282) ≈ 14.159 N i) The frictional force between the ladder and the ground (F_f). From Equation 1, F_f = N_A. F_f = 14.159 N Rounding to three significant figures: F_f = 14.2 N ii) The resultant force at the bottom of the ladder. The resultant force at the bottom (point B) is the vector sum of the normal force N_B and the frictional force F_f. Since these forces are perpendicular, we use the Pythagorean theorem. R_B = sqrt(N_B^2 + F_f^2) R_B = sqrt((98.1 N))^2 + (14.159 N)^2 R_B = sqrt(9623.61 + 200.477) = sqrt(9824.087) ≈ 99.116 N Rounding to three significant figures: R_B = 99.1 N iii) The coefficient of friction . Assuming the ladder is in limiting equilibrium (or we are finding the minimum coefficient of static friction required to prevent slipping), the frictional force is given by F_f = N_B. = (F_f)/(N_B) = 14.159 N98.1 N ≈ 0.14433 Rounding to three significant figures: = 0.144 iv) The angle of friction . The angle of friction is the angle whose tangent is the coefficient of friction. = = () = (0.14433) ≈ 8.209^ Rounding to one decimal place: = 8.2^ That's 2 down. 3 left today — send the next one.