Question

H Mz mi

The two masses "m1" and "m2" shown in the figure connected by a massless string and are being dropped by a constant horizontal force F a rough horizontal surface. F = 100 N, m1=10 kg, m2=15 kg coefficient kinetic friction between each mass and M_k= 0.2

expression: M2-->M1--> F

Questions:

1) Calculate the friction force on M2

2) Calculate the acceleration of the system of the 2 masses

3) Calculate the tension T in the string.

Answer 1

I mig -> T Ny (a) qen M (6) F=looN m. - lokg, m₂ = 15kg ME 0.2 Force equilibrium: Efyzo (figure a) N = m₂ g = 15x9.81 = 147.15 N Friction F= McN2 = 147.15x0.2 = 29.43 N b) Friction force on body-l: Fir unig = 0,24 10x9,81 F 19.62 N By force equilibrium along x for complete system: EF=ma F -FI-F2 = (mitm₂) a loo- 29.43 - 19.62 = (10+15)a a = 2.038 mls c) From (3): by force equilibrium: -T+F -Fi = ma. T=F-fi- ma = 100 - 19.62 - 10x 2.038 1 - 60N -