Question

this part. The system shown is at rest when a constant force E 50 lb is applied to collar B. Pre2 Next > 3 of 5

Answer 1

E = 50lb Consider the diagram on the right T^vector is the tension of the string F^vector is the Force on the string they are equal and opposite and magnitude is 6 pound Now, effective, Force on the collar is 50 - 6 = 44 pound towards left acceleration a = 44 lb. wt/18 = 44 times 32.2/18 ft/s^2 TildeEqual 78.7 ft/s^2 Now, we have formula v^2 = u^+ 2as here u = 0, a = 78.7 ft/s^2 s = 2ft v = squareroot 2as = squareroot 2 times 78.7 times 2 = 17.74ft/s Lets say the required distance be x then at x the velocity is squareroot 2a_1 x = squareroot 2 times 78.7 times x Now, if 50 mu_0 force is removed then the deceleration will be a_2 = 6 times 32.2/18 = 10.7ft/s^2 0^2 = u^2 - 2a_2 (2 - x) implies 2 times 78.2 times x = 2 times 10.7 times 2 - 2 times 10.7 times x