Presentation on theme: "Quiz * Vector A has a magnitude of 4 units and makes an angle of 37°"— Presentation transcript:
1 Quiz * Vector A has a magnitude of 4 units and makes an angle of 37° with the positive x-axis.Vector B has a magnitude of 6 units and makes an angle of 120°with the positive x- axis.Find the x-component and the y-component of each vector.Find A – B.
2 Two vectors are given by A = 3i + 3j , and B = -2i + 7j. Find 1. A + B,2. A – B,3. Find the angle between the two vectors
3 3.2.1The first condition for Equilibrium: For an object to be at rest, Newton's second law tells us that the sum of the forces acting on it must add up to zero. Since force is a vector, the components of the net force must each be zero. Hence, a condition for equilibrium is that∑ Fx = 0 , ∑Fy = 0 , ∑Fz =Equation 3.4 are called the First Condition for Equilibrium.
4 Example 3. 2: For chandelier cord tension Example 3.2: For chandelier cord tension. Calculate the tensions FA and FB in the two cords that are connected to the vertical cord supporting the 200kg chandelier in Fig 3.2.Solution:We first resolve FA into its horizontal (x) andvertical (y) components.Although we don’t know the value of FA.We can write,FAx = - FAcos60° and FAy=FAsin60°FB has only an x – component. In the vertical direction, we have the downward force exerted by the vertical cord equal to the weight of the chandelier, and the vertical component of FA upward.The weight of the chandelier isW = mg= 200 × 9.8 = 1960NSince, ∑Fy = 0 , we have∑Fy = , FA sin 60 – w = 0FA sin 60 – 1960 = 0
5 So, In the horizontal direction, ∑ Fx = FB – FA cos 60° = 0 FB = FA cos 60 = 2263 cos60 = 1131 N The magnitude of FA and FB determine the strength of cord or wire that must be used.
6 3.2.2 The Second Condition For Equilibrium The second condition for equilibrium: that the sum of the torques acting on an object, as calculated about any axis, must be zero:∑τ = (3.5)
7 Example 3. 3: For Balancing a seesaw. A board of mass M=2 Example 3.3: For Balancing a seesaw. A board of mass M=2.0kg serves as a seesaw for two children, as shown in Fig3.3a, child A has a mass of 30kg and sits 2.5m from the pivot point, P (his center of gravity is 2.5m from the pivot). At what distance x from the pivot must child B, of mass 25kg, place herself to balance the seesaw? Assume the board is uniform and centered over the pivot.Solution:We draw free body diagram, as shown Fig.3.5bkThe First: force equation. All the forces are in the y (vertical) direction, so∑Fy = 0 ,FN – mAg – mBg – Mg = 0Where FA = mAg and FB = mBg , because each child is in equilibrium when the seesaw is balanced.The Second: Torque equation is ∑τ = 0,mAg (2.5m) – mBg + Mg(0m) + FN (0m) = 0or,mAg (2.5m) – mBgx = 0thenTo balance the seesaw, child B must sit so that her CM is 3.0m from the pivot point.
8 Example 3. 4: For forces an a beam and supports Example 3.4: For forces an a beam and supports. A uniform 1500 kg beam, 20.0m long, supports a 15,000 kg printing press 5.0m from the right support column Fig Calculate the force on each of the vertical support columns.Solution:The torque , ∑τ = 0 , with the counterclockwise direction as positive gives∑τ = - (10.0m) (1500kg)g – (15.0m) (15,000kg)g + (20.0m)FB = 0Solving for FB , we find = (12,000 kg)g = 118,000N.To find FA, we use ∑Fy = 0 , with + y upward∑Fy = FA – (1500 kg)g – (15,000 kg)g + FB = 0Putting in FB = (12,000 kg)g, we find thatFA = (4500 kg)g = 44,100 N
9 The sum of the forces in the vertical (y) direction is ∑Fy = 0 , Example 3.5: For Hinged beam and cable. A uniform beam, 2.20m long with mass m = 25.0 kg, is mounted by a hinge on a wall as shown in Fig The beam is held in a horizontal position by a cable that makes an angle θ = 30.0° as shown. The beam supports a sign of mass M = 28.0 kg suspended from its end. Determine the components of the force FH that the hings exerts on the beam, and the tension FT in the supporting cable.Solution:The sum of the forces in the vertical (y) direction is∑Fy = 0 ,FHy + FTy – mg – Mg = (i)In the horizontal (x) direction, the sumof the forces is∑Fx = 0 ,FHx - FTx = (ii)For the torque equation, we choose the axis of the point where FT and Mg act. We choose torques that tend to rotate the beam counterclockwise as positive. The weight mg of the beam acts at its center, so we have
10 ∑τ = 0 - (FHy) (2.20 m) + mg (1.10m) = 0 We solve for FHy: Next, since the tension FT in the cable acts along the cable (θ = 30.0°). We see from Fig 3.7 that , or FTy = FTx tanθ = FTx tan30° = FTx (iv) Equation (i) above gives; FTy= (m+M)g – FHy = (53.0kg) (9.8m/s2) N = 396N Equation (iv) and (ii) give; FHx = FTx = 687N The components of FH are FHy = 123N and FHx = 687 N. the tension in the wire is
11 Example 3.5: For ladder. A 5.0m – long ladder leans a against a wall at a point 4.0m above a cement floor as shown in Fig The ladder is a uniform and has mass m= 12.0kg. Assuming the wall is frictionless (but the floor is not)k determine the forces exerted on the ladder by the floor and by the wall.We use the equilibrium conditions,∑Fx = , ∑Fy = , ∑τ = 0We will need all three since there are three unknowns:FW , Fcx, and Fcy .The y component of the force equation is∑Fy = Fcy - mg = 0So immediately we haveFcy = mg = 118NThe x component of the force equation is∑Fx = Fcx - FW = 0To determine both Fcx and FW, we need a torque equation. If we choose to calculate torques about an axis through the point where the ladder touches the cement floor, then Fc, which acts at this point, will have a lever arm of zero and so won’t enter the equation. The ladder touches the floor a distance;
12 The lever arm for mg is half this, or 1 The lever arm for mg is half this, or 1.5m, and the lever arm for FW is 4.0m, Fig3.7, we get ∑τ = (4.0m)FW – (1.5m)mg = 0 Thus, Then, from the x – component of the force equation, Fcx = FW = 44N Since the components of Fc are Fcx = 44N and Fcy = 118N, then And it acts at an angle to the floor of