Question

The collar of mass m = 3.7 kg slides on the rough horizontal shaft under the action of the force F of constant magnitude F = 24 N but variable direction. If θ = 0.63t, and if the collar has a speed v₁= 1.9 m/s to the right when θ = 0, determine the velocity v₂ of the collar when θ reaches 90°. Also determine the value of F which renders v₂ = v₁. The coefficient of friction is μ_k = 0.17.

F e т

Answer 1

Write the balanced force equation in vertical direction F sin eN W = mg N mg-Fsin e Write the balanced force equation in horizontal direction = ma Fcose-4 (mg-Fsine) F cos (mg - Fsin e) net m dv 1 [Fcose-mg +4Fsin e dt m [Fcos @ - м,mg +F sin @| dt dv S de Fcos -mg+4Fsin0 0.63 1 F cos-mg+^Fsine]de 0.63m Integrate the equation 1 -[F cos @ - д,mg+ u, F sin @de d J0.63m 1 [F sin 0-mge-4Fcose 0.63m (2- (1) 1 [Fsin e- mge-4Fcos 0,63m Substitute the given values

|-(0.17)(24) cos 90° 24) sin 90° -(0.17) (3.7) (9.81) 1 2(19)063(3.7)\ -{(24)sin 0° -(0.17) (3.7)(9.81) (0)- (0.17) (24) cos 0 -(0. (1)-(0.17)(3.7)(9.81) 1 =(1.9)+063(3.7)-{(24)-0-0-(0.17)(24)(1)} | (0.17)(3.7) (9.81) = (1.9)0.63(3.7)-{0-0-(0.17)(24)(1)} 1 + (0.1718-1081017)/24) 1 (1.9)+0.63(3.7) +(0.17) (24) (24)-(0.17)(3.7)(9.81) =9.788 m/s =0, from the equation (1). For v v-

1 (v-0.63m [F sin 0-4mge-44Fcose = 1 [F sin e-umge-44Fcose 0 0.63m F sin -mge-HFcos0 (0.17) (3.7)(9.81) 2 42mge F = sin 90°(0.17) cos 90° sin0 cos + (0.17)(3.7)(9.81) 2 1+0 =9.693 N