Equilibrium problems?

Question

if the centre of gravity of a hockey stick is at 60 cm from one end. to balance it at 40 cm from same end a weight of 5N is hung at the same end. what is the weight of the hockey stick?

ans= 10 N
Solution?

A painter weighing 700 N is standing on a platform which is hung bu two vertical ropes. the weight of the platform is 200 N and has 3m length. what will be the tension in the ropes if the painter is standing 1m away from one one?

Ans= 566.7 N and 333.3 N
i need solution.

Answer 1

For equilibrium the sum of the forces in the x and y directions must = zero. Additionally, the sum of the torques about ANY point must = 0.

For the hockey stick:
Pick your torque pt to be the end that's mentioned.

Now, since the sum of the y forces must be zero, the force acting upward at the balance pt must be w + 5
where w = weightforce of stick in newtons. Zero torque
abt the end mentioned implies:

(w + 5)40 - w(60) = 0 Solve for w.

For the painter, do choose the pivot pt as the center of his platform and t2 is the tension of the rope he's standing closer to.

t1 + t2 = 900 from the sum of the y forces must be zero.

so t2 = 900 - t1

That the torque abt any point is zero implies:

700(0.5) + t1(1.5) - (900 - t1)(1.5) = 0 Solve for t1; get t2 by subtracting t1 from 900.

Answer 2

u can solve the second problem by balncing torque at any point
(preferably centre of mass)
for a body to be in equilibrium net torque at any point must be zero
torque is defined as force X perpendicular distance of force from the point about which torque is taken

Answer 3

i dont feel tat d questions r clear...something is missing in both d problems...
plzz ckeck it..
by d way, draw d free body diagram and balance d forces in x,y,z directions..or balance d torques.. as at balanced condition d net torque will be zero .. ...
u will get equillibrium equations.. solve them to get d result..