QuizQuiz

* 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.

a)Find the x-component and the y-component of each vector.b)Find A – B.

Two vectors are given by Two vectors are given by AA = 3 = 3ii + 3 + 3jj , and , and BB = -2 = -2ii + 7 + 7jj. Find . Find

1. A1. A + + B,B, 2. A2. A – – BB,, 3. Find the angle between the two vectors3. Find the angle between the two vectors

3.2.1The first condition for Equilibrium3.2.1The first condition for Equilibrium::

For an object to be at rest, Newton's second law tells us that the 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 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. is a vector, the components of the net force must each be zero.

Hence, a condition for equilibrium is thatHence, a condition for equilibrium is that

∑∑ Fx = 0 , Fx = 0 , ∑∑Fy = 0 , Fy = 0 , ∑∑Fz = 0 3.4Fz = 0 3.4

Equation 3.4 are called the First Condition for EquilibriumEquation 3.4 are called the First Condition for Equilibrium . .

Example 3.2Example 3.2:: For chandelier cord tension. Calculate the For chandelier cord tension. Calculate the tensions tensions FA FA and and FBFB in the two cords that are connected to the in the two cords that are connected to the vertical cord supporting the 200kg chandelier in Fig 3.2.vertical cord supporting the 200kg chandelier in Fig 3.2.

Solution:Solution:We first resolve We first resolve FAFA into its horizontal (x) and into its horizontal (x) and vertical (y) components. vertical (y) components. Although we don’t know the value of Although we don’t know the value of FAFA..We can write, We can write, FAx = - FAcos60° and FAy=FAsin60°FAx = - FAcos60° and FAy=FAsin60°

FBFB has only an x – component. In the vertical direction, we have the downward 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 force exerted by the vertical cord equal to the weight of the chandelier, and the vertical component of FA upward.vertical component of FA upward.The weight of the chandelier is The weight of the chandelier is W = mgW = mg = 200= 200 × 9.8 = 1960N× 9.8 = 1960NSince, Since, ∑∑Fy = 0 , we have Fy = 0 , we have ∑∑Fy = , FA sin 60 – w = 0 Fy = , FA sin 60 – w = 0 FA sin 60 – 1960 = 0FA sin 60 – 1960 = 0

So, So, In the horizontal direction, In the horizontal direction, ∑∑ Fx = FB – FA cos 60° = 0 Fx = FB – FA cos 60° = 0FB = FA cos 60 = 2263 cos60 = 1131 NFB = FA cos 60 = 2263 cos60 = 1131 NThe magnitude of The magnitude of FAFA and and FBFB determine the strength of cord or determine the strength of cord or wire that must be usedwire that must be used..

3.2.2 3.2.2 The Second Condition For EquilibriumThe Second Condition For Equilibrium

The second condition for equilibriumThe second condition for equilibrium: that the : that the sum of the torques acting on an object, as sum of the torques acting on an object, as calculated about any axis, must be zero: calculated about any axis, must be zero: ∑ ∑τ = 0 (3.5)τ = 0 (3.5)

Example 3.3Example 3.3:: For Balancing a seesaw. A board of mass M=2.0kg serves as a 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 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 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 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 pivotseesaw? Assume the board is uniform and centered over the pivot..

SolutionSolution:: We draw free body diagram, as shown Fig.3.5bkWe draw free body diagram, as shown Fig.3.5bkThe First: force equation. All the forces are in the y (vertical) direction, soThe First: force equation. All the forces are in the y (vertical) direction, so

∑∑Fy = 0Fy = 0, , FN – mAg – mBg – Mg = 0FN – mAg – mBg – Mg = 0Where FA = mAg and FB = mBg , because each child is in equilibrium when the Where FA = mAg and FB = mBg , because each child is in equilibrium when the

seesaw is balanced.seesaw is balanced.The Second: Torque equation is The Second: Torque equation is ∑∑τ = 0,τ = 0,mAg (2.5m) – mBg + Mg(0m) + FN (0m) = 0mAg (2.5m) – mBg + Mg(0m) + FN (0m) = 0or, or, mAg (2.5m) – mBgx = 0mAg (2.5m) – mBgx = 0then then To balance the seesaw, child B must sit so that her CM is 3.0m from the pivot To balance the seesaw, child B must sit so that her CM is 3.0m from the pivot

point.point.

mm

mx

B

A 0.3)5.2(25

30)5.2(

Example 3.4Example 3.4:: For forces an a beam and supports. A uniform 1500 For forces an a beam and supports. A uniform 1500 kg beam, 20.0m long, supports a 15,000 kg printing press 5.0m kg beam, 20.0m long, supports a 15,000 kg printing press 5.0m from the right support column Fig. 3.6. Calculate the force on each from the right support column Fig. 3.6. Calculate the force on each

of the vertical support columnsof the vertical support columns.. SolutionSolution::

The torque , The torque , ∑∑τ = 0 , with the counterclockwise direction as positive τ = 0 , with the counterclockwise direction as positive givesgives

∑∑τ = - (10.0m) (1500kg)g – (15.0m) (15,000kg)g + (20.0m)FB = 0τ = - (10.0m) (1500kg)g – (15.0m) (15,000kg)g + (20.0m)FB = 0Solving for Solving for FBFB , we find = (12,000 kg)g = 118,000N , we find = (12,000 kg)g = 118,000N . .

To find To find FAFA, we use , we use ∑∑Fy = 0 , with + y upwardFy = 0 , with + y upward

∑∑FyFy = = FAFA – (1500 kg)g – (15,000 kg)g + – (1500 kg)g – (15,000 kg)g + FBFB = 0 = 0Putting in FB = (12,000 kg)g, we find thatPutting in FB = (12,000 kg)g, we find that FA = (4500 kg)g = 44,100 NFA = (4500 kg)g = 44,100 N

Example 3.5Example 3.5: For Hinged beam and cable. A uniform beam, 2.20m long with : 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 3.7. The mass m = 25.0 kg, is mounted by a hinge on a wall as shown in Fig 3.7. The beam is held in a horizontal position by a cable that makes an angle θ = 30.0° as 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. shown. The beam supports a sign of mass M = 28.0 kg suspended from its end. Determine the components of the force Determine the components of the force FHFH that the hings exerts on the beam, and that the hings exerts on the beam, and the tension the tension FTFT in the supporting cable in the supporting cable..

Solution: Solution: The sum of the forces in the vertical (y) direction is The sum of the forces in the vertical (y) direction is ∑ ∑Fy = 0 ,Fy = 0 ,FHy + FTy – mg – Mg = 0 (i)FHy + FTy – mg – Mg = 0 (i)In the horizontal (x) direction, the sum In the horizontal (x) direction, the sum of the forces is of the forces is ∑ ∑Fx = 0 ,Fx = 0 , FHx - FTx = 0 (ii)FHx - FTx = 0 (ii)For the torque equation, we choose the axis of the point where For the torque equation, we choose the axis of the point where FTFT

and Mand Mgg act. We choose torques that tend to rotate the beam act. We choose torques that tend to rotate the beam counterclockwise as positive. The weight mg of the beam acts at counterclockwise as positive. The weight mg of the beam acts at its center, so we haveits center, so we have

∑∑τ = 0 τ = 0 - (FHy) (2.20 m) + mg (1.10m) = 0- (FHy) (2.20 m) + mg (1.10m) = 0We solve for FHy:We solve for FHy:

Next, since the tension Next, since the tension FTFT in the cable acts along the cable ( in the cable acts along the cable (θθ = 30.0°). = 30.0°). We see from Fig 3.7 that , or We see from Fig 3.7 that , or FTy = FTx tanFTy = FTx tanθθ = FTx tan30° = 0.577 FTx (iv) = FTx tan30° = 0.577 FTx (iv)Equation (i) above gives; Equation (i) above gives; FTy= (m+M)g – FHy = (53.0kg) (9.8m/s2) - 133N = 396NFTy= (m+M)g – FHy = (53.0kg) (9.8m/s2) - 133N = 396NEquation (iv) and (ii) give;Equation (iv) and (ii) give;

FHx = FTx = 687NFHx = FTx = 687N

The components of FH are FHy = 123N and FHx = 687 N. the tension in the The components of FH are FHy = 123N and FHx = 687 N. the tension in the

wire iswire is

Nsmkgmgm

mFHy 133)/80.9)(0.25)(500.0(

20.2

10.1 2

NF

F TyTx 687

577.0

NFFF TyTxT 79322

Example 3.5:Example 3.5: For ladder. A 5.0m – long ladder leans a against a wall at a point 4.0m above a 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 3.7. The ladder is a uniform and has mass m= cement floor as shown in Fig 3.7. The ladder is a uniform and has mass m= 12.0kg. Assuming the wall is frictionless (but the floor is not)k determine the 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 wallforces exerted on the ladder by the floor and by the wall.. We use the equilibrium conditionsWe use the equilibrium conditions , ,

∑∑Fx = 0 , Fx = 0 , ∑∑Fy = 0 , Fy = 0 , ∑∑τ = 0τ = 0We will need all three since there are three unknownsWe will need all three since there are three unknowns : :FW , FcxFW , Fcx, and , and Fcy Fcy ..

The y component of the force equation isThe y component of the force equation is ∑∑Fy Fy == Fcy Fcy - mg = 0- mg = 0

So immediately we haveSo immediately we have Fcy Fcy = mg = 118N= mg = 118NThe x component of the force equation isThe x component of the force equation is

∑∑Fx Fx == Fcx Fcx - - FWFW = 0 = 0To determine both Fcx and FW, we need a torque equation. If we choose to To 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 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 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 distanceand so won’t enter the equation. The ladder touches the floor a distance;;

mmmx 0.3)0.4()0.5( 220

The lever arm for mg is half this, or 1.5m, and the lever arm for FW is 4.0m, The lever arm for mg is half this, or 1.5m, and the lever arm for FW is 4.0m, Fig3.7, we getFig3.7, we get∑∑τ τ = (4.0m)FW – (1.5m)mg = 0= (4.0m)FW – (1.5m)mg = 0

ThusThus,, Then, from the x – component of the force equation, Then, from the x – component of the force equation, Fcx = FW = 44NFcx = FW = 44NSince the components of Fc are Fcx = 44N and Fcy = 118N, thenSince the components of Fc are Fcx = 44N and Fcy = 118N, then

And it acts at an angle to the floor ofAnd it acts at an angle to the floor of

Nm

smkgmFW 44

0.4

)/8.9)(0.12)(5.1( 2

NNNNFc 130126)118()44( 22

7044

118tan 1

N

N