Preview File Edit View Go Tools Window Help Q Thu Oct 12 7:51 PM Worksheet 5 - F23 Page 1 of 6 Q Q 1 6 - 5 6 . Q Search Worksheet 5 - F23 A. Forces, Decomposition of Forces and Free-body Diagrams 1) A force is a "push or pull" acting on an object. It has magnitude, direction. Its tail is always at the acting 380.0 N point. Since it is vector, we can use the following formulas: a. Fx = F cos 0 Fy = F sin 0. b. F = VF2+ F2 (R / Fx ) 1 30 Example 1: Two forces F, and F2 acting on a car F2 have magnitudes and angles as shown in the picture on the right. Therefore, their components are: 450.0 N Fix = F1 cos 01 = 380.0 cos 10.00 = 374.2 (N) Fly = F1 sin 01 = 380.0 sin 10.00 = 66.0 (N) F2x = F2 COS 02 = 450.0 cos -30.0 = 389.7 (N) 2 Fzy = F2 sin 02 = 450.0 sin -30.00 = -225.0 (N) 2) A free-body diagram is a sketch representing an object as a point-like particle with all the external forces acting on it. There are some particular forces including: gravitational force Fg, normal force FN, friction force f, tension force T, elastic force Felastic etc. System 1 Free-body diagrams System 2 N Fnoor 3 Fear System 1 Forof N' Froot Fnoor O System 26 Preview File Edit View 60 Tools Window Help Q E (63) >3 E} 6? Q 8 C ThuOct'lZ 7:52PM V Worksheet 5 - F23 a] Page 'l of 6 2) A free-body diagram is a sketch representing an object as a point-like particle with all the external forces acting on it. There are some particular forces including: gravitational force 171,, normal force 17'\Preview File Edit View Go Tools Window Help Q Thu Oct 12 7:52 PM Worksheet 5 - F23 Page 2 of 6 1 Q Q 1 2 . ' A Q search Worksheet 5 - F23 B. Net force and 3 Newton's laws: () Net force is the resultant vector of all forces on an object. Its action is equivalent to all forces when replacing them. Object Object Object In vector form: Fnet = [F = F1+ F2+ F3 + Fat ... In component form: Fnet,x = Fix + Fax + Fax + Fax + ... Fnet,y = Fly + Fzy + Fay + Fay + ... 2 Example 3: For the car in the Example 1, calculate its net force components, magnitude and angle. Fnet,x = Fix + Fzx = 374.2 + 389.7 = 763.9 (N) Fnet,y = Fly + Fzy = 66.0 - 225.0 = -159.0 (N) Fnet = Fret,x + Fret,y = V763.92 + (-159.0)2 = 780.3 (N) 0 = tan-1 (Fnet.y /Fnetx) = tan-1(-159.0/763.9) = -11.60 3 2) Newton's First Law: If no net force acting on an object, the object velocity can not change or its acceleration is equal to zero. Fnet = 0 = v = const = a = 0 3) Newton's Second Law: A net force on an object is equal to the product of the object's mass and its acceleration. In vector form: Fnet = maPreview File Edit View Go Tools Window Help Q Thu Oct 12 7:52 PM Worksheet 5 - F23 Page 2 of 6 1 Q Q 1 2 ~ D' A . Q search Worksheet 5 - F23 Fnet = 0 = v = const = a = 0 3) Newton's Second Law: A net force on an object is equal to the product of the object's mass and its acceleration. In vector form: Fnet = ma In component form: Fnet,x = max Fnet,y = may Example 4: If the car in the Example I has a mass of 1200 kg, calculate its acceleration components, magnitude and angle. Fnet,x = max = ax = Fnet,x/m = 763.9/1200 = 0.6366 (32) Fnet,y = may = dy = Fnet,y/m = 763.-159.0/1200 = -0.1325 (-2) anet = Jaz + a3 = 0.63662 + (-0.1325)2 = 0.6502 (52) 0 = tan-1 (dy/ax) = tan-1(-0.1325/0.6366) = -11.60 2 4) Newton's Third Law: When two objects interact, the forces on the objects from each other are always equal in magnitude and opposite in direction. They are called action - reaction. Note that they act on different objects. 3Preview File Edit View Go Tools Window Help Q Thu Oct 12 7:52 PM Worksheet 5 - F23 1 Q Q " 2 . ' A . Q search Page 3 of 6 Worksheet 5 - F23 C. Problems: I) Three forces act on an object as in the figure. F, has a magnitude YA F1 of 4.00 N and angle of 60.0. F2 and F3 have magnitude of 2.00 N and 5.00 N, respectively. Determine: a) Each force components X F3 2 b) Net force components, magnitude and angle 3Preview File Edit View Go Tools Window Help Q Thu Oct 12 7:53 PM Worksheet 5 - F23 Page 4 of 6 1 Q Q 1 2 v ' A Q search Worksheet 5 - F23 c) If the object has a mass of 2 (kg), what is its acceleration components, magnitude and angle? 2 2) A person drags a 30.0 kg box across the floor by pulling on a rope at an angle of 0 = 20.0 above the horizontal. 3 a) Sketch the free-body for the system considering the box and the rope as one object. 4Preview File Edit View Go Tools Window Help Q Thu Oct 12 7:53 PM Worksheet 5 - F23 Page 4 of 6 Q Q 1 2 - 5 6 . Q Search Worksheet 5 - F23 2) A person drags a 30.0 kg box across the floor by pulling on a rope at an angle of 0 = 20.0 above the horizontal. a) Sketch the free-body for the system considering the box and the rope as one object. 2 b) Suppose the force of gravity on the box is 300 N, the horizontal friction force is 120 N, and the 3 force of the person pulling on the box is 160 N. Use Newton's second law to determine any unknown force(s) and component(s) of the acceleration. 46 Preview File Edit View 60 Tools Window Help Q E (r?) >3 E} A Q 8 C ThuOct'lZ 7:53PM Worksheet 5 - F23 a] V Page 5 of 6 c) Assuming the acceleration you calculated is constant, how much time does it take for the person to pull the box a distance of 10 m if the box starts from rest? 3) A force is applied to a block to move it up a 300 incline. The incline is frictionless If F = 65.0 N and M : 5.00 kg, what is the magnitude of the acceleration of the block? WM\" '9 Wicca" I. +1334!\