One end of a meter stick is pinned to a table, so the stick can rotate freely in

Question

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 2.20 N and is applied perpendicular to the length of the stick at the free end. The other force has a magnitude of 5.60 N and acts at a 32.0° angle with respect to the length of the stick. Where along the stick is the 5.60 N force applied? Express this distance with respect to the end that is pinned.
______________m

Explanation / Answer

if we take counter clock wise torques as positiv e then the force of first force    1 = F1 L1         = 2.2N* 1m = 2.2Nm the torque due to the second force 2 = - F2 sin600 L2                                                            =- 5.6 N sin 32 L2                                                          = - 2.96N L2 since net torque is zero    = 0       2.2Nm + ( -2.96 N) L2 = 0        2.2Nm = 2.96 N L2 L2 = 2.2Nm/ 2.96N L2 =0.7432 m L2 = 2.2Nm/ 2.96N L2 =0.7432 m

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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