tutorial the laws of motion0

8/10/2019 Tutorial the Laws of Motion0

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PHYS101 The Laws of Motion Fall 2012/2013

The Laws of Motion1. An object of mass m1 = 55.00 kg placed on a frictionless, horizontal table is

connected to a string that passes over a pulley and then is fastened to a hangingobject of massm2=59.00 kg as shown in Figure 1.

(a) Draw free-body diagrams of both objects.

(b) Find the magnitude of the acceleration of the objects

(c) Find the tension in the string.

(d) If there is a frictionk=0.2 on the surface, repeat the part a ,b, c, d

Solutions:

(a) Free-Body Diagrams (Notice the chosen directions for acceleration)

(b) Applying the second law form1

Fx =T=m1a (1)

Fy =n

m1g= 0 (2)form2

Fx =0 (3)

c2013 Department of Physics, Eastern Mediterranean University Page 1 of 13


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Fy =m2g T= m2a (4)from (1) and (4)

m2gm1a= m2a

m2g= (m1+m2)a

a= m2

(m1+m2)g (5)

m1=55kg andm2=59kg and g = 9.81ms2

a= 59kg

59kg+55kg 9.81ms2 =5.08ms2

(c) To find the tension we may use (1)

T= m1a= 55kg 5.08ms2 =279.4N

(d) Free-Body diagram when friction exists between the surfacesform1

Fx =T fk=m1a ; fk=kn (6)

Fy = m1g n= 0 n= m1g (7)

from (6) and (7)T km1g= m1a (8)

form2Fx =0 (9)

Fy =m2g T= m2a (10)

Now we add (8) and (10)

m2g km1g= (m1+m2)a

a= (m2 km1)m1+m2

g (11)

a=(59kg 0.2 55kg) 9.81ms2

55kg+59kg =4.13ms2

and for tension we may use (10)

T= m2(g a) =59kg (9.81ms2 4.13ms2) =335.12N



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(e)

2. Two objects are connected by a light string that passes over a frictionless pulleyas shown in Figure P2. Assume the incline is frictionless and takem1 = 2.00 kg,m2 = 6.00 kg, and =55

(a) Draw free-body diagrams of both objects.

(b) Find the magnitude of the acceleration of the objects

(c) Find the tension in the string.

(d) Find the speed of each object 2.00 s after it is released from rest.

(e) If there is a frictionk=0.2 on the surface, repeat the part a ,b, c, d

Solutions:



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(a) Free-Body Diagram (Notice the chosen directions for acceleration)

(b) Apply the second law form1

Fx =0 (12)

Fy =Tm1g= m1a (13)

form2 Fx =m2g sin T= m2a (14)

Fy =n m2g cos = 0 (15)

Adding (13) and (14) we get

m2g sin m1g= (m1+m2)a

a=(m2sin m1)

m1+m2g (16)

a=(6kg sin55 2kg)

2kg+6kg 9.81ms2 =3.57ms2

(c) To calculate tension we use (13)

T= m1(g+a) =2kg (9.81ms2 +3.57ms2) =26.8N

(d) For both objects we havevi = 0 and we may usevf =vi+at

vf =at = 3.57ms2 2.00s= 7.14ms1

(e) Free-Body Diagram:

We now apply the second law tom1andm2when the incline is frictionless.form1

Fx =0 (17)



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Fy =Tm1g= m1a (18)and form2

Fx =m2g sin T fk=m2g ; fk=kn (19)

Fy =n m2g cos =0 n= m2g cos (20)

from (19) and (20)

m2g sin T km2g cos = m2a (21)

Adding (18) and (21) results in

m2g sin km2g cos m1g= (m1+m2)a

a=m2(sin kcos )m1

m1+m2g (22)

a=6kg(sin55 0.2 cos55) 2kg

2kg+6kg 9.81ms2 =2.73ms2

from (18)

T=m1(g+a) =2kg (9.81ms2

+2.73ms2

) =25.08N

Its speed att = 2.00sis

vf =at = 2.73ms2 2.00s= 5.46ms1

(f)



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3. A woman at an airport is pulling her 20.0-kg suitcase at constant speed bypulling on a strap at an angle above the horizontal. She pulls on the strap witha 35.0-N force, and the friction force on the suitcase is 20.0N.

(a) Draw a freebody diagram of the suitcase.

(b) What angle does the strap make with the horizontal?

(c) What is the magnitude of the normal force that the ground exerts on the

suitcase?Solutions:

(a) Free-Body Diagram

(b) Apply the second law to the suitcase but note that the right-hand side ofthe equation on the horizontal must be zero since the suitcase is moved atconstant speed.

Fx =F cos fk=0 cos = fk

F

(23)

cos =20.0N

35.0N =0.57 = arccos (0.57) =55.15



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(c) Write the second law on the vertical

Fy =n+F sin mg = 0 (24)

n= (20.0kg)(9.81ms2) 35.0N sin (55.15) =167.5N

4. Two blocks connected by a rope of negligible mass are being dragged by a hor-izontal force (Fig. P4). Suppose F = 68.0 N, m1 = 12 kg, m2 = 18 kg andthe coefficient of kinetic friction between each block and the surface is 0.100 .

(a) Draw a free-body diagram for each block.

(b) Determine the acceleration of the system.

(c) Determine the tension T in the rope.

Solutions:

(a) Free-Body Diagram

(b) Apply the second law to each block separately.Form1

Fx =T fk1 =m1a ; fk1 =kn1 (25)

Fy =n1 m1g= 0 n1=m1g (26)

from (25) and (26) we have

T km1g= m1a (27)

form2Fx =F T fk2 =m2a ; fk2 =kn2 (28)



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Fy =n2 m2g= 0 n2=m2g (29)from (28) and (29)

F T km2g= m2a (30)

We now add (27) and (30) to obtain

F kg(m1+m2) = (m1+m2)a

a= F kg(m1+m2)

m1+m2(31)

a=68.0N (0.100)(9.81ms2)(12.0kg+18.0kg)

12.0kg+18.0kg =1.29ms2

(c) From (27)

T= (kg+a)m1 = [(0.100)(9.81ms2) +1.29ms2] 12.0kg = 27.3N

5. In Fig. 5, a tin of antioxidants(m1 = 1.0kg) on a frictionless inclined surfaceis connected to a tin of corned beef(m2 = 2.0kg). The pulley is massless andfrictionless. An upward force of magnitudeF = 6.0Nacts on the corned beeftin, which has a downward acceleration of 5.5m/s2 .

(a) What is the tension in the connecting cord ?

(b) What is the angle ?

Solutions:



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(a) To work out tension we apply the second law tom2.

Fx =0 (32)

Fy =m2g T F= m2a T= (g a)m2 F (33)

T= (9.81ms2 5.5ms2)(2.0kg) 6.0N= 2.62N

(b) To find the angle apply the second law tom1.

Fx =T+m1g sin= m1a sin=m1a T

m1g

(34)

sin=(1.0kg)(5.5ms2) 2.62N

1.0kg 9.81ms2 =0.3

= Arcsin(0.3) =17.1 (35)

Note that applying the second law for the vertical direction gives us infor-mation which is of no benefit for this particular problem.

6. In Figure P6, the pulleys and the cord are light, all surfaces are frictionless, andthe cord does not stretch.



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(a) How does the acceleration of block 1 compare with the acceleration of block2?

(b) The mass of block 2 is 1.30 kg. Find its acceleration as it depends on themassm1of block 1. Evaluate your answer form1 = 0.550kg.

(c) What If? What does the result of part (b) predict ifm1 is very much lessthan 1.30 kg?

(d) What does the result of part (b) predict ifm1approaches infinity?

(e) In this last case, what is the tension in the long cord?

(f) Could you anticipate the answers to parts (c), (d), and (e) without first do-

ing part (b)? Explain.

Solutions:

(a) The moving pulley has an accelerationa2. Since m1 moves twice the dis-tance this pulley moves in the same time, m1 has twice the acceleration ofthe pulley. Hencea1=2a2

(b)m2g T2=m2a2 (36)

T1 = m1a1 (37)

T2 2T1 =0 (38)

from (36), (37), and (38) and the fact thata1=2a2

a2= m2m2+4m1

g (39)

a2 = 1.30kg

1.30kg+4 0.550kg 9.81ms2 =3.64ms2 (40)

(c) Ifm1is very much less thanm2 =1.30kg then the term in (39) includingm1is negligible compared tom2and hence the acceleration ofm2approachesg= 9.81ms2.



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(d) Ifm1approaches infinity in (39)a2will approach zero.(e) Ifa2approaches zero in (36) T2 = m2gand from (38)T1 =

12m2g.

(f) In case (c) since m1 is negligible compared tom2 we can simply concludethatm2 freely falls with g. In (d) an infinitely massive block, m1, does notlet the other block move so it is predictable that the accelerationa2is zero.And finally in case (e) since the block 2 is stationary the forces ,weight andtension, acting on it must balance and knowing the fact that T2 = 2T1 thesame conclusion as we did in (e) comes out and that isT1=

12m2g.

7. What horizontal force must be applied to a large block of mass M shown in Fig-

ure 7 so that the tan blocks remain stationary relative to M? Assume all surfacesand the pulley are frictionless. Notice that the force exerted by the string acceler-

ates m2.Solutions:

F= (M+m1+m2)a (41)form1

T= m2a (42)



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form2 Tm1g= 0 (43)

Eliminating T we get

m2am1g= 0 a=m1m2

g (44)

for the entire system

F= (M+m1+m2)(m1m2

)g (45)

8. A car accelerates down a hill (Fig. P 8), going from rest to 30.0m/s in 6.00 s. A toyinside the car hangs by a string from the cars ceiling. The ball in the figure rep-resents the toy, of mass 0.100 kg. The acceleration is such that the string remains

perpendicular to the ceiling.

(a) Determine the angle?

(b) Determine the tension in the string?

Solutions:



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(a) Choosing x axis pointing down the slope

vf =vi+at

30.0ms1 =0+a(6.00ms2) a= 5.00ms2

Fx =mg sin =ma sin = a

g (46)

sin =5.00ms2

9.81ms2 =0.51 = Arcsin(0.51) =30.7

(b)Fy =Tmg cos = 0 (47)

T= (0.100kg)(9.81ms2) cos (30.7) =0.843N

9. The two blocks(m= 16.0kg)and(M= 88.0kg)in Fig. 9 are not attached to eachother.The coefficient of static friction between the blocks is k = 0.38, but thesurface beneath the larger block is frictionless. What is the minimum magnitudeof the horizontal force required to keep the smaller block from slipping down the

larger block?

Solutions:

F= (m+M)a a= F

m+M (48)

for m

Fy =m

(49)Fx =F n= ma (50)



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from (43), (44), and (45)F=

(m+M)mg

Ms(51)

F=(16.0kg+88.0kg) 16.0kg 9.81ms2

88.0kg 0.38 =488N


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