Question

During a slam-dunk, a basketball player pulls on the hoop with a 250 lbf at point C of the ring as shown in the figure . Find the moment of the force about a) the point of the ring attachment to the board (point B), and b) the root of the pole, point O.

Answer 1

Solution: It has been given that the force applied the force applied is 250 lbf at 15° to the downwards direction. It can be represented as: Fsinis" Bott A A Also می ممF F its rectilinear components Upon breaking the given force into we get Fcod 1505 (250 lbf) (0.966) 241.5 lbf. Esin 15º = (250 lbf) (0.259) - 64.75 db6. : کو ů moment of force is defined as: Moment of Force Perpendicular laka Torque) Distance :) (Force) At point B, only the vertical component is able to generate a moment of torque. Hence moment of force at point B is given as: Perpen dicular distance 241.5 lbf) between B and Force B (115ft) (24105 Abb) 362.25 dbf-ft Feos 15 Lo (in clockwise direction) ). el.51 B .

we have (i) similarly, for point o Moment of force due to the horizontal component 2 perpendicular distance between o and (Fsin 15°) F sin 150 force force (lo bo ) (64.75 lbf) 647.5 lbf-ft anticlockwiseo al B -Esino 15° to perpendicular distance between horizontal component of force and point o.

Moment of force due to vertical component is: 020 Perpendicular distance Feas 159 between o and force Ecos15° force 0) = 은 14.540)/241.5268) clockwise . = = 1086.75 dbf-ft. 4.5', L 3' 3 115 sis FCOS15° : perpendicular distance between vertical component and point o .: Net moment of force about point o és : 647.5 dbf-ft antid ockwise +(003 1086.75 dbf-ft clockwise (9086.75- 647.5) lbf-ft clockwise ) ol 439.25 lbf-ft clockwise