Question

A long wooden post is suspended horizontally by a rope attached to the ceiling at distance Δx = 1.20 m from the end. A pylon supports the other end as shown below. The 4.85 m long post is uniform along its length and has a mass of 35.5-kg. A 10.0-kg mass is suspended at a distance Δx = 1.20 m from the other end.

-Ax— AX pylon 10 kg

(a) Find the tension in the rope.
N

(b) Find the force that the pylon exerts on the end of the post.

Answer 1

Givem the diatance $\Delta$ x = 1.2m, the length of the post L = 4.85m, mass of the post M = 35.5kg and the mass suspended m = 10kg. The forces acting on the system is shown below.

AT * k- AM-- 1M --- -4M- 3 M

(a) For the system to be in equilibrium, the net torque acting on the system shouls be zero. Let T be the tension in the rope.Taking torque about the pylon, we get

$mg\times \Delta x+Mg\times \frac{L}{2}-T\times (L-\Delta x)=0$

mg x 1.c + Mg x=TⓇ (L - A.c)

10 x 9.8 x 1.2 + 35.5 x 9.8 x 4.85 2 =T * (4.85 – 1.2)

$117.6+843.66=T\times 3.65$

$961.26=T\times 3.65$

$T=\frac{961.26}{3.65}=263.36N$

So the tension in the rope is 363.36N.

(b) Similarly, taking torque about the point O, we get,

$Mg\times (\frac{L}{2}-\Delta x)+mg\times (L-2\Delta x)-N\times (L-\Delta x)=0$

$Mg\times (\frac{L}{2}-\Delta x)+mg\times (L-2\Delta x)=N\times (L-\Delta x)$

$35.5\times 9.8\times (\frac{4.85}{2}-1.2)+10\times 9.8\times (4.85-2\times 1.2)=N\times (4.85-1.2)$

$35.5\times 9.8\times 1.225+10\times 9.8\times 2.45=N\times 3.65$

$426.18+240.1=N\times 3.65$

$666.28=N\times 3.65$

$N=\frac{666.28}{3.65}=182.54N$

So the force that the pylon exerts on the end of the post is 182.54N.