Question

Mmm to B A collar B of mass m is attached to the spring AB and can move along the rod shown. The constant of the spring is k = 1.5 kN/m and the spring is unstretched (neutral) when E=0°. Knowing that the coefficient of static friction between the collar and the rod is 0.40, determine the range of values of m for which equilibrium can be maintained. Please show work and include a free body diagram. Please explain as much as you can. I want to learn this material and not just have the solution. Thank you! b= 0.5 m 0= 30°

Answer 1

Case When collar tries to move upward, frictional force 'acts in downward direction Here is the free body diagram AN Fs = spring force fs=static friction N= normal force m = mass of collor fs t mg

Coso = x= sehen Extension in the spring, Ax = x-b AX = Cabo -b Ax = b (telo-1) sx =(0.57 (Totsó-) Ax = 0.0774 m String force = Fs = K AX

Fr= [1.5(1000) N/m] (0.0774m) Fs = 116.02 N Component of force in horizontal direction 'Fsx = Es coso =116-02N) (Cosbó) = 100.5 N, Component of force along y axis, Fsy = Fs Sino = (116.02N) Sin30) = 58.015 N Since, the collor is at rest, so, net force in all directions would be zero. net force along horizontal direction, N- Fsx =O

N = Fsx = -100.5N Maximum static frict on = fs = MSN Lfs = 0.4)(100.5N) fs = 40.21 Now, net force in vertical direction, Fsy - mg - fs =0 58.015N -mg – 40.2N =D m= 17.8151 - 17.8150 m= 9.8m/s2 m= 1.82 kg Ansel

Now, motion of the collor tends to downward, fraction would act upward and operoyaning foot y A 4 IS diagram would be happen. In this case, het force in vertical direction=o Fsy tfs-ng=o 58.015N + 40.2N – M (9.8m152=0

98.25N m= 9.8m/s² m= 10.02kg] Thus, The range of mass varies from 1.82 kg to 10.02 kg. Ans,