Trapdoor Problem

balance the moment of tension and weight about the hinge. t(9/15)(12)=(50)(6)

3T/5 * 1.2 = 50 * 0.6 .......... 6T = 250 ........... T = 41.67 N

equate the vertical components of the torque due to the thread and the weight of doortrap about the point of hinge...

T 3/5 120=50*60

force balance due to Tension "T" in the string and the weight "W", 1st we calculate the angle made by string from vertical,i.e, tan(theta)=90/120 =36.87 degree from horizontal or 90-36.87=53.13 degree from vertical. Therefore, Now take moment about hinge to be zero; (Tcos (theta)]120}-{50*(weight=60)=0; =41.67 N

Its a Basic Problem from elementary mechanics,i.e, Torque balancing or you may call it a simple force balance due to Tension "T" in the string and the weight "W", Therefore 1st we have to calculate the angle made by string from vertical,i.e, tan(theta)=90/120 => theta=tan^-1(3/4)=36.87 degree from horizontal or 90-36.87=53.13 degree from vertical. Therefore, Now take moment about hinge to be zero; =>{ [T cos(theta)] 120}-{50*(weight=60)}=0; =>T=41.67 N

to lift the trapdoor ..which rotates about the Hinge ..we need to overcome the roating effect due to WEIGHT..about the hinge... so balnce the torque due to tension..... with torque due to weight about hinge...THAT will BE just enough force to lift it up

considering the moment about O ..Tsin theta * 1.2 = mg * 0.6 mg = 50N and sin theta = 3/5.. on solving this equation we get T= 41.67N

Consider the rotational equilibrium about point O.

Take the moment at the hinge , but before you do that find the angle by tan@= 90/120, @= 36.87, then find the moments about the hinge to find reaction

Who knows how To find reaction at hinge