Question

Consider the ladder in 8.15. Compute the normal force again, assuming a 75 kg man is standing on the ladder (a) at its midpoint and (b) four fifths of the way up the latter.

I posted this question earlier but was very confused on how it was answered and laid out. I am not understanding how to go about this or what formulas to use.

EXAMPLE 8.15 Leaning Ladder A ladder leans against a wall, at 16 to the vertical. It has length m. However, you can apply rotational equilibrium, summing the L = 3.64 m and mass m 18.2 kg. There's no friction at the wall. but there is friction at the floor. Find the normal force of the wall on the ladder. torques about the bottom of the ladder. Then the two forces acting on that point have r = 0, so don't contribute to the torque. This leaves a positive torque due to the wall force and a negative torque due to the ladder's weight. They sum to zero in equilibrium. ORGANIZE AND PLAN At the outset, it's not clear whether we should use translational equilibrium, rotational equilibrium, or both. There are four forces acting on the ladder: its weight w = ng, the normal force from L/ 2 irom the bottom. with the weight vector at 16 to the radius at that the wall w, the normal force from the floor , and static friction a point. The wall force acts at distance L from the bottom and makes an the ground, which keeps the ladder from slipping sideways (Figure 8.2) l901674 with the radius. Therefore, the net torque is SOLVE Assuming the ladder is uniform, its weight acts at its center Of these, we're given only the ladder's weight. With so many un- known forces, translational equilibrium alone won't solve the prob- Tjer =-(L/2)(mg)(sin 16") + (L)(%)(sin 74°) = 0 Solving for the wall's normal force gives mg sin 16 2 sin 74 18.2 kg) (980 m/s-sn 6 2 sin 74 25.6 N nw REFLECT Notice how changing the angle affects the answer. Make il smaller, and the normal force decreases reaching zero for a vertical ladder. But make the angle larger, and the normal force increases- requiring a correspondingly larger frictional force for equilibrium; see "Making the Connection." below m = 18.2 kg 16 16 MAKING THE CONNECTION Based on the answer in this example, find the normal and frictional forces on the floor. mg ANSWER With weight mg 178 N being the only other vertical force the floor's normal force must also be 178 N. Meanwhile, the frictional force must point horizontally to the right, with magnitude 25.6 N, to balance the wall's normal force. If the frictional coefficient is too small, μ,mg won't be that big, and then the ladder slips. EIGURE 8.21 Forces acting on the ladder

I know the answers is (a) 130.95 N and (b) 194.2 N but I do not know how to get them. THank you

**ALL DIAGRAMS ARE HERE*

Answer 1

Sigma T = 0 -(L/2)(M + m)g sin 16 degree + L N_W sin 74 = 0 N_W = (M + m)g sin 16/2 sin (74) = (75 + 18.2) times 9.81 times sin 16/2 sin 74 N_W = 131 N -(L/2)mg sin 16 - (4L/5) mg sin 16 + L N_W sin (90 - 16) = 0 - 1/2 times 18.2 times 9.8 times sin 16 - 4/5 times 75 times 9.8 times sin 16 + N_W sin 74 = 0 N_W = 186.66/sin 74 N_W = 194.2 N