A uniform horizontal rod of mass 1.2 kg and length 0.83 m is free to pivot about

Question

A uniform horizontal rod of mass 1.2 kg and length 0.83 m is free to pivot about one end as shown. The moment of inertia of the rod about an perpendicular to the rod and through the center of mass is given in the equation sheet. If 5.1 N force at an angle of 120, to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? Given information F = 10 N, m_1 = 8.5 kg, m_2 = 2.4 kg, m_3 = 1.25 kg, R = 0.12 m, and mu = 0.223. Find the acceleration of the system, and Find T_1 and T_2.

Explanation / Answer

PROBLEM 4:for the case of mass1:

T1=m1*g

m1=8.5kg

T1=8.5*10=85N

for the case of mass2:

two forces act towards the left side of the mass2 i.e..FX AND FF

ONE FORCE TOWARDS THE RIGHT SIDE i.e FY,force due to mass1.

(A)1 st linear :

sum of the forces= FY+(-FX)+(-FF)=m*a

FX=m2*g*sin 20=24*0.342=8.2084N

FY=m1*g=85N

FF=0.223*m2*g*cos 20

=0.223*24*0.9396

=5.029N

85-8.2084-5.029=(2.4+8.5)*a

a=71.7626/10.9

a=6.5837 m/sec^2

(B)to find tension T2:

T2 +(-FX)+(-FY)=(m1+m2)*a

T2-8.2084-5.029=10.9*6.5837

T2=84.999N

T1=m1*g

T1=8.5*10=85N

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