Problem 9-3: The bar AB of mass m 10 kg and length L = 1 m is released from rest

Question

Problem 9-3: The bar AB of mass m 10 kg and length L = 1 m is released from rest on the plane surfaces shown in the diagram with the angle =30°. The vertical surface is frictionless, while the coefficients of static and kinetic friction between the horizontal surface and the bar are 0.6 and Ax=0.5, respectively. Determine if the bar is in motion or not. If it is in motion, then determine the angular acceleration of the bar, and the surface normal forces applied on the bar at its endpoints A and B. Smooth A

Explanation / Answer

normal force at B is nB

sum of the vertical forces is zero

nB-(m*g) = 0

nB = m*g = 10*9.81 = 98.1

maximum static frictional force = fs(max) = mu_s*nB = 0.6*98.1 = 58.86 N

normal force at A is nA = fs(max) = 58.86 N

kinetic frictional force is mu_k*nB = 0.5*98.1 = 49.05 N

Net torque acting on system is

(nA*1*sin(30))+(m*g*0.5*sin(30))

(58.86*1*sin(30)) - (10*9.81*0.5*cos(30)) = 13 which is not equal to zero

hence the bar is not rest it will be in motion

2) we know that torque is T = I*alpha

I is the moment of inertia = (1/12)*m*l^2 + (3m/16) = (1/12)*10*1^2 + (3*10/16) = 2.7 kg-m^2

alpha is the angular accelaration

but T = (fk*1*sin(30))-(m*g*0.5*sin(30)) = 2.7*alpha

(49.05*1*sin(30)) - (10*9.81*0.5*cos(30)) = 2.7*alpha

alpha = 6.65 rad/s^2

3) reaction at A is nA = fr =49.05 N

and reaction at B is nB = 10*9.81 = 98.1 N

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