# A uniform beam of mass MB = 0.

A uniform beam of mass MB = 0.8833 kg and length l = 0.8m isattached to a wall by a hinge at at one end and by a cord ofnegligible mass and length ls = 1.27 m that makes an angle ? withthe horizontal at the other end (see figure drawn on the board),such that the beam is horizontal. A mass mH = 0.55 kg is hung fromthe beam at a distance lH = 0.6 m from the hinge. What is thetension in the cord, and what is the magnitude and direction of theforce between the hinge and beam?

Answer:

**a) angle made by cord with the beam, theta =cos^-1(0.8/1.27)**

**= 50.96 degrees**

**As the beam is in equilibrium net force and net torqueacting on the beam must be zero.**

**let T is the tension in the cord.**

**net torque about hinge = 0**

**0.8333*9.8*(0.8/2) + 0.55*9.8*0.6*sin(90) -T*0.8*sin(50.96) = 0**

**T = (0.8333*9.8*(0.8/2) +0.55*9.8*0.6*sin(90))/(0.8*sin(50.96))**

**= 10.5 N<<<<<<<————-Answer**

**b) let Fx and Fy are component of force exerted by thehinge.**

**use, Fnetx = 0**

**Fx – T*cos(50.96) = 0**

**Fx = T*cos(50.96)**

**= 10.5*cos(50.96)**

**= 6.61 N**

**Apply, Fnety = 0**

**Fy + T*sin(50.96) – 0.8833*9.8 – 0.55*9.8 =0**

**Fy = 0.8833*9.8 + 0.55*9.8 -10.5*sin(50.96)**

**= 5.89 N**

**F_hinge = sqrt(Fx^2 +Fy^2)**

**= sqrt(6.61^2 + 5.89^2)**

**= 8.85 N<<<<<<<<<<———————-Answer**

**direction : theta = tan^-1(Fy/Fx)**

**= tan^-1(5.89/6.61)**

**= 41.7 degrees above +x axis<<<<<<<<<<———————-Answer**