Question

The figure shows a model of a crane that may be mounted on a truck. A rigid uniform horizontal bar of mass shows a model of a crane that may be mounted on a truck. A rigid uniform horizontal bar of mass m_1 = 95.00 kg and length L = 6.000 m is supported by two vertical massless strings. String A is attached at a distance d = 1.300 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass m_2 = 3000 kg is supported by the crane at a distance x = 5.800 m from the left end of the bar. Throughout this problem, positive torque is counterclockwise. Use 9.81 m/s for the magnitude of the acceleration due to gravity.

1. Find T_A, the tension in string A.

2. Find T_B, the magnitude of the tension in string B

Answer 1

take moments about left end

T_A*d -m1*g*L/2 - m2*g*x = 0

T_A = 133453.73 N

taking moments about point A gives

T_B *d - m1*g*(l/2 - d) - m2*g*(x-d) = 0

solving T_B = 103091.78 N

Answer 2

1.Sum torque about string B
Ta*d-m1*g*L/2-m2*g*x=0
solve for Ta

Ta=g*(m1*L/2+m2*x)/d
plug in the numbers
Ta=9.81*(80*5.3/2+3500*5.1)/1
177188.22 N

for Tb, sum forces in the y
177188.22-Tb-g*3580=0
solve for Tb

Tb=177188.22-9.81*3580
142068.42 N
__________________-