Question

A cord acb, 5 m long is attached at points a and b to two vertical walls 3 m apart as shown in fig-2.A pull c of negligible radius carries a suspended load of 200 n and is free to roll without friction along the cord. Determine the position of equilibrium, as defined by the distance x, that the pulley will assume and also the tensile force in the cord.

Answer 1

Answer:

Explanation:

=> The pulley C is in equilibrium under the action of tensile forces in CA and CB and suspended load 200 N. The tensile forces in segment CA and CB are equal as the pulley is frictionless. let's consider the equilibrium of pulley C.

ΣH = 0

T cos θ_1 - T cos θ_2 = 0

θ_1 = θ_2 [that equal to θ ]

=> Suppose, BC be extended to D

Thus, ΔCFD = ΔCFA

∴ CD = AC

BD = BC + CD = BC + AC = length of chord = 5m

DE = 3 m  

BE = 4 m

=> Here, ∆BHI is similar to ∆BDE

HI = BI / BE * DE

= 1/4 * 3  

= 3/4

= 0.75

=> Thus, AH = 3 - 0.75

∴ AH= 2.25

=> As,  AH = 2x

x = 2.25 / 2  

x = 1.125 m

=> At point C, ΣV = 0

2 * T sin ፀ = 200

2 * T * 4/5 = 200

T = 1000 / 8

= 125 N