Solution: (a)The second law force diagrams of two mass are shown in figure. Set the coordinate system as in pply Newton,'s second law of motion, we have re Assuming the acceleration is a shown in fig m1g m2g N-m,g cos 8=0 m2-m, sing 2.6-95×sin34 =-22m/s2 m1+m2 5+ m1m28 9.5×2.6×sin34 (+sin 8) (1+sin34)=31 9.5+26 (b The second law force diagrams of two mass are shown in figure. Set the coordinate system as in Assuming the acceleration is a shown in figure. N Apply Newton's second law of motion, we have m,gsin 8-T-f=m,a m1g g N-m,g cos0=0 m,g=m2 f=urN Solving the equations, we get n, ( sin Ak 9.5×(sin34+0.15c0s34)-26 =1.2m/s m,+ 9.5+26 仍+m(1+sinO-C0:95×26×sin34 m,m2g 95+)×(1+sin34-0.15×cos34)=29N 3. An object is drop from rest. Find the terminal speed assuming that the drag D force is given by D= by Solution The forces acting on the object and the coordinate system are shown in figureSolution: (a) The second law force diagrams of two mass are shown in figure. Set the coordinate system as in figure. Assuming the acceleration is a v shown in figure. Apply Newton’s second law of motion, we have ⎪ ⎩ ⎪ ⎨ ⎧ − = − = − = m g T m a N m g T m g m a 2 2 1 1 1 cos 0 sin θ θ Solving it, we get 2 1 2 2 1 2.2 m/s 9.5 2.6 sin 2.6 9.5 sin 34 = − + − × = + − = o g m m m m a θ (1 sin 34 ) 31 N 9.5 2.6 9.5 2.6 sin 34 (1 sin ) 1 2 1 2 × + = + × × + = + = o o θ m m m m g T (b) The second law force diagrams of two mass are shown in figure. Set the coordinate system as in figure. Assuming the acceleration is a v shown in figure. Apply Newton’s second law of motion, we have ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = − − = f N T m g m a N m g m g T f m a µ k θ θ 2 2 1 1 1 cos 0 sin Solving the equations, we get 2 1 2 1 2 1.2 m/s 9.5 2.6 (sin cos ) 9.5 (sin 34 0.15cos34 ) 2.6 = + × + − = + − − = o o g m m m m a θ µ k θ (1 sin34 0.15 cos34 ) 29 N 9.5 2.6 9.5 2.6 sin34 (1 sin cos ) 1 2 1 2 × + − × = + × × + − = + = o o o θ µk θ m m m m g T 3. An object is drop from rest. Find the terminal speed assuming that the drag force is given by 2 D = bv . Solution: The forces acting on the object and the coordinate system are shown in figure. Apply Newton’s second law of motion, we have mg v D v j ˆ m1 m2 x y m g v 1 T v m g v 2 N v T v a v m1 m2 x y m g v 1 T v m g v 2 N v T v a v f v a v