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

dang that balls went far

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