A long, uniform beam with mass m and length L is attached by means of a pivot, l

Question

A long, uniform beam with mass m and length L is attached by means of a pivot, located at L/4, to a vertical support as shown above. The beam is connected to a support line oriented at an angle 0 relative to the horizontal. The beam is in equilibrium in the horizontal position when a weight of mass 2m is placed at the left end of the beam and a mass of 3m is placed at the right end of the beam. Calculate the tension Tin the support line. Calculate the x and y components of force acting on the pivot point.

Explanation / Answer

a)

here

this is a static equilibrium problem where the sum of the torques acting on the beam is zero

(L/4) 2*m*g - (L/4) m* g - (3L /4) * 3 m *g + (3L/4) T *sin(theta) = 0

T = 8mg / 3sin(theta)

b)

to determine the Rx and Ry components we will use newton's second law in horizontal and vertical directions

Horizontally :

Fx = 0

Rx - Tx = 0

Rx = Tx = T cos(theta) = 8mg /3tan(theta)

vertically :

Fy = 0

Ry + Ty - F2mg - Fg - F3mg = 0

Ry = 2mg + mg + 3mg - Tsin(theta)

Ry = 6mg - (8mg / 3sin(theta) ) sin(theta) = 10 mg / 3

c)

I = Ibeam + M2 * R2^2 + M3R3^2

I = 7 * ML^2 / 48 + 2M(L/4)^2 + 3M(3L/4)^2

I = 94 ML^2 / 48

then

torque = I * alpha

L * 2mg / 4 - L *mg / 4 - 3L * 3mg /4 = (94 ML^2 / 48) alpha

alpha = -96 g / 94L = -48g / 47L

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