Question

a 90kg mail bag hangs by a vertical rope 3.5m ling. a postal worker then displaces the bag to a position 2m sideways from its original position, always keeping the rope taut. what horizontal force is necessary to hold the bag in the new position? as the bag is moved to this position, how much work is done by the rope

Answer 1

$$F=mg\cot{\theta}=mg\frac{\sqrt{l^2-d^2}}{d}$$

The horizontal force :

$$F=(90)(9.8)\frac{\sqrt{3.5^2-2^2}}{2}=1300\ N$$

The work done by the rope:

$$W=mgh=mg\sqrt{l^2-d^2}=(90)(9.8)\sqrt{3.5^2-2^2}=2500\ J$$