Consider the following situation 0.3 =0.2 =30° A 2 kg box on an inclined table.
ID: 1779025 • Letter: C
Question
Consider the following situation 0.3 =0.2 =30° A 2 kg box on an inclined table. Please follow the instructions for this homework carefully. This problem and the next three pertain to the same physical situation. List the forces acting on the box (we will consider these forces acting on the box as the "action" forces): Force #1 Force #2 Force #3 The net force acting on the box has magnitude of and is directed The acceleration of the box has magnitude of and is directed What would the magnitude of this box's acceleration if the table top was inclined at the 45o to horizontal instead of 30? What would the magnitude of this box's acceleration if the table top was inclined at the 60% to horizontal instead of 30°?Explanation / Answer
Force 1 : gravitation force
Force 2 : static friction
Force 3 : kinetic friction
Gravitation force
Fg = mg = 2*9.8 = 19.6 N
Gravitation force in inclined direction
Fgi = Fg*cos(60°) = 9.8 N
Force perpendicular to inclined -
N=Fgn = Fg*sin(60°) = 16.97 N
Static friction -
Fs = us*N =0.3*16.97 =5.09 N
Kinetic friction
Fk =uk*N =0.2*16.97= 3.4N
Net force
Fnet = Fgi - Fk = 9.8-3.4 = 6.4 N
Direction: downwards inclined (30° downwards)
Fnet = ma
Net acceleration
Anet = 6.4/2= 3.2 m/s2
Direction: downwards inclined (30° downwards)
If inclined to 45°
Fgi = 13.86 N
N=Fgn = 13 .86N
Fs =4.16 N
Fk=2.77 N
Anet= (13.86-2.77)/2 = 5.5 m/s2
If inclined to 60°
Fgi = 16.97N
Fgn= 9.8N
Fk=1.96 N
Anet = (16.97-1.96)/2 = 7.5 m/s2