A 16.3 kg person climbs up a uniform 137 N

ladder. The upper and lower ends of the ladder rest on frictionless surfaces. The bottom

of the ladder is fastened to the wall by a horizontal rope that can support a maximum tension of 116 N. The angle between the horizontal and the ladder is 52◦.

A) Find the tension in the rope when the person is one-third of the way up the ladder. The acceleration of gravity is 9.8 m/s2. Answer in units of N

B) Find the maximum distance dmax the person can climb up the ladder before the rope

breaks. Answer in units of m

A) T = 95.12 N, B) L_man - 0.59 L

Explanation:

A) For this problem we must use the equilibrium, translational and rotational equations

∑τ = 0

Let's fix our reference system at the top of the stairs, the anti-clockwise rotation is positive

W x/2 + W_man x_man - T y = 0 (1)

Let's look with trigonometry for distances

cos 52 = x / L

x = L cos 52

sin 52 = y / L

y = L sin52

The man indicates it is at L / 3, so its distance is

x_man = L/3 cos 52

We replace

W L/2 cos 52 + W_man L/3 cos 52 - T L sin 52 = 0

T = cos 52 (W / 2 + W_man / 3) / sin 52

T = cos 52 (137/2 + 16.3 9.8 / 3) / sin 52

T = 121.75 ctan 52

T = 95.12 N

B) let's look for the maximum distance (x) that can be before breaking the rope

L_man = (-W L / 2 cos 52 + T L sin 52) / W_man cos 52

L_man / L = (-W / 2 ctan 52 + T tan 52) / W_man

L_man / L = (-137/2 ctan 52 + 116 tan 52) / (16.3 9.8)

L_man / L = (-53.518 + 148.47) / 159.74

L_ma / L = 0.594

The man then climb a length of 0,59 (59%) along the length of the stairs