The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A

Question

The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass = 100.0kg and length = 5.700m is supported by two vertical massless strings. String A is attached at a distance = 1.700m 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 = 3500kg is supported by the crane at a distance = 5.500m from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.807m/s^2 for the magnitude of the acceleration due to gravity. A. Find , the tension in string A. (In N) B. Find , the magnitude of the tension in string B. (In N)

Explanation / Answer

As bar is in rotational equilibrium,taking torque at left end Torque due to T(B) =0 Torque due to T(A) =T(A)*1.4 counterclockwise Torque due to weight of bar = 80*9.81*3=2354.4 N clockwise Torque due to weight of object =3500*9.81*5.8=199143 N clockwise counterclockwise torque =clockwise torque T(A)*1.4 =201497.4 T(A) =143926.71 N T(B) + m1g +m2g =T(A) T(B)=143926.71 -3580*9.81 T(B) =108806.91 N (1) T(A) the tension in string A.is 143926.71 N (2) T(B) the magnitude of the tension in string B is 108806.91 N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.