Question

A local automobile dealership has chosen to hang a car outside the entry to their showroom. The car has been modified so that its mass is 645 kg. The car's weight will be supported by a heavy-duty chain which pulls upwards and leftwards. A beam will push outward to balance the wallwardpull of the chain. The manufacturer of the chain claims it has a breaking strength of 17400 Newtons. Village code requires Chain that there be a safety factor of 2.20. That is, the actual tension in the chain must be no more than 1/2.20 of the breaking strength. What is the minimum angle (between the chain and the horizontal) at which the dealership can hang the car? 1) What has to be the case for the object to hang in equilibrium? 2) What does the safety factor affect? 3) Draw a free body diagram. 4) What are the x and y components of each of the forces? 5) What is the sum of the x components and sum of the y components of the forces shown? 6) What is the minimum angle between the chain and the beam?

Answer 1

3 free body dia goam. m= 645 kg A (Tchain) allow Tenais = 17400 N f = 2.20, og foram - - (Tchain) allow - Tchains - 17400 f 2.2 V W twang = (Tchain) allow-=.7909.09 N (1) cases of equilibrium. Efy = 0 Efx=0 (2) Factor of safety affects the maximum weight of - the car which can be hanged. (4) [(Ichamakonde 7909.09 cosa [(Tchain allow) g +7909.9 smo Jawowy W= -645*9.81 N , Freams " 5) For, equilibrium Efyão as 2909.g sino-64589.81=0 Ex=0 = -7909.09 coso + Fream=0 (6) solving the above equations Sino645X981 2909.9 © = 53-139] Ang = 53,13