Question

2. The 7-kg collar slides with negligible friction on the fired vertical shaft When the collar is relessed from rest at the bottom posyon shown it moves up the shaft under the action of the constant force F= 200 450 mm N applied to the cable Calculate the stiffness which the spring must have if its maximum compression is limited to 75 mm The position of the small pulley at Bis fixed

Answer 1

solution- Draw the F.B.1) for given Diagram. Fara 75mm 225mm att usornm motion final position In the FoBD A is New position of collon when external force is applied Initial length of cable As is AB= 0.uso+(0.22512 TAB = 0.5031m Fincal length of cable A'B is A B = AB- distance ko.Pulley & shaft - 0.503) -0.225 A'R = 0.3781m Now work done by the collar due to external : Force is {W = F XAB) DJ W = 200 X 0.2781 wa 55.62 nim CamScanner Scanned with!

.. From Wook energy principle, ne = mgh it a ka? Given that m=768, 9=9817/02 n-) heisht moved by collar =) h= 0.450m a manimum compression of sping > x=0.075m ky spring stiffness . - W = mght kn2 (5562)= (7*9-81X0.45) #&*K70.0752) 55.62 = 30.90 +0.00281**) -> IK = 8.797 kN lm to the stiffness of the spring is lik= 8.797 kn/m] .. Scanned with Camcanner