Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 06

Find the magnitude of acceleration of the two masses and the tension in the cord consider m1 = 4 kg and m2= 6 kg.. we show the car’s weight →W acting downwards, the normal force → N acti

In part (c) of Fig.5.11, we first construct a free-body diagram of the stationary knot that holds the three cords together, and then we choose the coordinate axes

By applying Newton’s second law in the x and y directions of part (c) of the figure,

we find that:

F x = T3cosθ − T2cosφ = 0 and F y = T3sinθ + T2sinφ − T1= 0

From the x-component equation we get the following relation:

T3= cosφ cosθ T2= 12/13

3/5 T2=20

13 T2

When we substitute the result of T3 into the y component equation, after putting

T1= mg = 210 N, we get:

20

13

4

5 T2+ 5

13T2− 210 = 0 ⇒

 16

13 + 5 13



T2= 210 ⇒ T2= 130 N

Consequently, one can find the value of the third tension to be:

T3=20

13 T2= 20

13× 130 N = 200 N

Example 5.3

Two masses m1 and m2 (m2 > m1), are connected by a light cord that passes over a

massless, frictionless pulley as shown in part (a) of Fig.5.12 This arrangement is called Atwood’s machine and sometimes is used to measure the acceleration due

to gravity Find the magnitude of acceleration of the two masses and the tension

in the cord (consider m1 = 4 kg and m2= 6 kg).

Solution: We construct a free-body diagram for the two masses as shown in parts

(b) and (c) of Fig.5.12 When Newton’s second law is applied to m1in part (b) of the figure, we find:

F y = T − m1g = m1a

Also, we do the same for m2of part (c) of the figure, to get:

(a) (b)

m1

m2

T

m1g

T

m2 g

a a

(c)

y

y

Fig 5.12

When we add the last two equations, T will cancel out, and we get:

m2g − m1g = m2a + m1a

Thus: a= m2− m1

m1+ m2

6 kg+ 4 kg× 9.8 m/s2= 1.96 m/s2

If we substitute with a into the first equation we get:

T = 2m1 m2

m1+ m2

6 kg+ 4 kg × 9.8 m/s2= 47 N

Example 5.4

A car moving with an initial speedv◦= 30 m/s suddenly brakes, locking its wheels

(i.e it starts to skid) The car travels on the road a distance d= 75 m before it comes

to a complete stop Find the coefficient of kinetic friction between the tires of the car and the road

Solution: Part (a) of Fig.5.13depicts the car’s travel In part (b) we choose the coordinate axes and show the car’s free-body diagram during its skid In this part

we show the car’s weight →

W (acting downwards), the normal force →

N (acting perpendicularly to the road), and kinetic frictional force →

fk(acting to the left)

x

y N

mg

fk

a

d

Fig 5.13

Applying Newton’s second law in the component form, we find that:

F x = − fk= ma

F y = N − mg = 0 From the last equation we get N = mg Since fk= μkN = μ k mg , then the first

equation gives:

−μkmg = ma

The negative sign means that the acceleration is to the left Since a is constant,

we can usev2= v2

◦+ 2ax, with v = 0 and x = d This gives:

0= v2

◦+ 2ad = v2

◦− 2μkgd

Thus: μk= v2◦

2gd = (30 m/s)2

2(9.8 m/s2)(75m) = 0.61

Example 5.5

A block of mass m = 2 kg is placed on an inclined plane of angle θ = 30◦, as

shown in Fig.5.14 The block is released from rest at the top of the plane, where the

distance from the bottom is d = 10 m (a) Find the magnitude of the acceleration

of the block and the normal force exerted on the block (b) How long does it take the block to reach the bottom, and what is its speed just as it gets there?

Solution: (a) We construct a free-body diagram to this example as shown in part

(b) of Fig.5.14 The only forces on the block are the weight →

W (acting downward) and the normal force →

N (acting perpendicular to the inclined plane) We choose a

coordinate system with x axis parallel to the incline and y-axis perpendicular to it.

With this choice, the angle between the weight vector and the negative direction

of the y-axis equals the angle θ of the inclined plane After that, we decompose the weight to a component of magnitude mg cos θ along the negative y-axis and

a component of magnitude mg sin θ along the x axis The block will slide along the inclined plane with acceleration ax and will never leave the plane; i.e ay = 0 Applying Newton’s second law to the x and y components gives:

F x = mg sin θ = m a x

F y = N − mg cos θ = 0

(a)

N

W=mg

(b)

θ

θ

x y

θ

θ

m g

sin

m g

cos

a x d

Fig 5.14

From the x-component form, we see that the acceleration along the incline is provided by the component of the weight down the incline By taking g= 10 m/s2,

we get:

a x = g sin θ = (10 m/s2)(sin 30◦) = 5 m/s2

Notice that whenθ = 0 (i.e when the plane is horizontal) we have a x= 0 (the minimum acceleration value) Also, we see that when θ = 90◦ (i.e when the

plane is vertical) the case resembles a free fall scenario, resulting in ax = g

(the maximum acceleration value)

From the y component of Newton’s second law we find N to be:

N = mg cos θ = (2 kg)(10 m/s2)(cos 30◦) = 17.32 N

Also, notice that whenθ = 0, we have N = mg = 20 N (its maximum value),

and whenθ = 90◦, we have N = 0 (its minimum value).

(b) Since ax = constant, we apply the equation x = v x◦t+1

2a x t2to the block withv x◦= 0 and x = d = 10 m to get the following relation:

d= 1

2a x t 2

Then, Solving for t and taking the positive root yields:

t=

2d

a x =



2× (10 m)

5 m/s2 = 2 s Also, we can apply the kinematics equationv2

x = v2

x◦+ 2a x x with v x◦= 0 and

x = d = 10 m to get the following relation:

v2

x = 2a x d

Then, solving forv xand taking the positive root yields:

v x= 2ax d=

2(5 m/s2)(10 m) = 10 m/s

Example 5.6

A block of mass m1= 4 kg lying on a rough horizontal surface is connected to a

second block of mass m2= 6 kg by a light non-stretchable cord over a massless, frictionless pulley as shown in part (a) of Fig.5.15 The coefficient of kinetic friction between the block and the surface isμk = 0.5 (a) Find the magnitudes

of the acceleration of the system and the tension in the cord (b) Find the relation

between m1 and m2in the case when the system is on the verge of slipping

(a) (c)

m1

m 2

(b)

T

m 1 g

a

y

T

y

m 2

m1

x

fk

N

x

Fig 5.15

Solution: (a) Since the cord is non-stretchable, the two masses have the same

magnitude of acceleration Consequently, we construct a free-body diagram for the two masses as shown in parts (b) and (c) of Fig.5.15, where we take the x axis

always along any of the body’s motion In this case a cannot take negative values When Newton’s second law is applied to m2in part (b) of the figure, we find:

(1) F x = m2g − T = m2a

F y= 0

From (1), we can find the magnitude of the tension in terms of g and a That is:

(2) T = m2g − m2a

Doing the same for m1(see part (c) of Fig.5.15) we get:

(3) F x = T − fk= m1a (4) F y = N − m1g= 0

Since fk = μkN , and from (4) we have N = m1g , then:

(5) fk= μkm1g

When this result is substituted into (3), we get:

(6) T = μkm1g + m1a

Equating the magnitude of the tension in (2) and (6), we get:

μ m g + m a = m (g − a)

Solving for a we get:

a= m2 − μkm1

m1+ m2

g

Note that, when m2 > μkm1we have accelerated motion, and when m2 = μkm1,

we have motion with zero acceleration, i.e the speed is constant The value of a

can then be evaluated as follows:

a= 6 kg− 0.5(4 kg)

6 kg+ 4 kg × 9.8 m/s2= 3.92 m/s2

We can find T by substituting the expression of a into (6), to get:

T = (μk+ 1) m1m2

m1+ m2

g

Thus: T =(0.5 + 1)(4 kg)(6 kg)

6 kg+ 4 kg × 9.8 m/s

2= 35.28 N

(b) When the system is on the verge of slipping, the magnitude of the force T that acts on mass m1 must equal the maximum static friction fs ,max = μsN , i.e.

T = μsN = μsm1g Also, the weight of the mass m2must equal the magnitude

of the tension, i.e T = m2g Thus:

m2g = μsm1g

Finally:

m2= μsm1 (On the verge of slipping)

Example 5.7

A block is at rest on a rough inclined plane of angle θ, as shown in Fig.5.16

(a) Find the static frictional force fs in terms of N and θ (b) When the angle

is increased until the block is on the verge of slipping atθ = θc= 38.7◦, find

the value of the coefficient of static frictionμs (c) After we increase θ further to

allow the block to accelerate and then decreaseθ again to the value θ = θ= 26.6◦

to allow the block to move with constant speed, find the coefficient of kinetic frictionμk.

y

θ

m g

cos

m g

sin

θ

fs

Solution: (a) The block is balanced under its weight mg, the normal force N, and

the static frictional force fs Taking x parallel to the plane and y perpendicular to

it, then Newton’s second law will give:

F x = mg sin θ − fs = 0

F y = N − mg cos θ = 0 From the last equation we find mg = N/ cos θ Therefore, we can eliminate

mg from the first equation to get:

fs = mg sin θ = N

cosθsinθ = N tan θ

(b) When the inclined plane is at the critical angleθc, the block is on the verge

of slipping and fs = fs,max = μsN So, at this angle the last equation becomes

μsN = N tan θc.

Thus: μs = tan θc −−−−−−−−−→

whenθc=38.7◦ μs = tan 38.7◦= 0.8

(c) When the block moves with constant speed atθ= 26.6◦, the kinetic friction

fk= μkN equals the weight component mg sin θ.

Thus:

whenθ=26.6◦ μk= tan 26.6◦= 0.5

Example 5.8

A small sphere of mass m = 1.5 g is released from rest in a large vessel filled

with liquid, see Fig.5.17 The sphere reaches a terminal speed ofvt= 2.45 cm/s.

Assume that the resistive drag force is given by Eq.5.11.∗(a) Solve Eq.5.12to

find the speed of the sphere as a function of time (b) Find the time t it takes the

sphere to reach a speed of 0.9 vt.

Fig 5.17

FD

t

y

m g

Solution:∗(a) To solve Eq.5.12, we setτ = m/b, which is called the time

con-stant, and perform the following steps:

mg − bv = m d v

0

d v

vt− v = τ−1

t

 0

dt

The previous integration can be performed to get:

v = vt(1 − e −t/τ )

One can find from this result that the timeτ = m/b is the time it takes the

sphere to reach 63% of its terminal speed

(b) Let us first determine the coefficient b in Eq.5.11 Since the terminal speed

is given by the relation FD = bvt = mg, i.e vt = mg/b, then the value of b will

be given by:

b=mg v

t =(1.5 × 10−3kg)(9.8 m/s2)

2.45 × 10−2m/s = 0.6 kg/s

Therefore, the value of the timeτ is given by:

τ = m

b = 1.5 × 10−3kg

0.6 kg/s = 2.5 × 10−3s= 2.5 ms

We set v = 0.9 vt in the resulting formula of part (a), and we perform the

following steps, to find the corresponding time t:

0.9 = 1 − e −t/τ ⇒ e −t/τ = 0.1 ⇒ t = −τ ln(0.1)

Thus: t = −(2.5 × 10−3s)(−2.303) = 5.76 × 10−3s= 5.76 ms

Section 5.3 Applications to Newton’s Laws

(Take g= 10 m/s2

in all the following exercises unless g is given)

(1) A 10 g bullet accelerates from rest to 500 m/s in a gun barrel of length 10 cm, see Fig.5.18 Find the accelerating force (assuming it constant)

(2) A horizontal cable pulls a golf cart of mass 400 kg along a horizontal track As

in Fig.5.19, the tension in the cable is 800 N (a) Starting from rest, how long will it take the cart to reach a speed of 10 m/s? (b) Find the distance covered during this time

Fig 5.19 See Exercise (2)

800 N

(3) A block of mass 2 kg is accelerated by the two forces→

F1 = 8→i + 3→j and

→

F2= −5→i − 7→j , (all units in newtons) (a) What is the net force on the block

in unit vector notation, and what is its magnitude and direction? (b) What is the magnitude and direction of the acceleration?

(4) Assume only five forces are acting in the xy plane on a block of mass 4 kg as

shown in Fig.5.20 (a) Taking sinθ = 4/5 and cos θ = 3/5, find the block’s

acceleration in unit-vector notation (b) Find the acceleration’s magnitude and direction

Fig 5.20 See Exercise (4)

5N

θ

x

15N

10N 1N

(5) Consider a block of weight W hanging from three ropes as shown in Fig.5.21 (a) At what angleθ will the magnitude of the tensions T2and T3each be equal

to the weight W (b) Is this angle independent of W?

Fig 5.21 See Exercise (5)

θ

T 2 T 3

T 1 W

θ

(6) A traffic light of mass m= 10 kg is suspended over a road as shown in Fig.5.22 The ropes are connected to the top of two vertical and identical posts at an angle

θ = 65◦ Find the magnitude of the tension in all three cables.

(7) A block of mass 20 kg is suspended from the ceiling and the wall by three cords tied together as shown in Fig.5.23 Find the magnitude of the tensions

T1, T2, and T3in the ropes

(8) After applying its brakes on a dry road, a 1,000 kg car moving atv◦= 40 m/s

requires a minimum distance d of 50 m to come to a complete stop without

skidding, see Fig.5.24 (a) Find the car’s acceleration (b) Find the frictional force exerted on the car by the road (c) Find the coefficient of static friction

Fig 5.22 See Exercise (6)

θ

m

T1

Fig 5.23 See Exercise (7)

70ο

60ο

T2

T3

T1 W

Fig 5.24 See Exercise (8)

d

a

(9) A small sphere of mass m is attached to one end of a massless thread The

other end of that thread is fixed in the roof of a truck when it is at rest

Take g = 9.8 m/s2 (a) What angle θ does the thread make with the verti-cal when the truck has a constant acceleration a = 1.5 m/s2, see Fig.5.25? (b) Findθ when the truck is moving with a constant velocity of magnitude

v = 100 km/h?

Fig 5.25 See Exercise (9)

m θ

a

(10) A small object is hanging by a thread from the rearview mirror of a sports car The car accelerates uniformly from rest to 90 km/h in 10 s What angle θ does

the thread make with the vertical?

(11) Two blocks of masses m1 = 4 kg and m2= 12 kg are in contact on a smooth

horizontal surface A horizontal force of magnitude F= 12 N pushes them as shown in Fig.5.26 (a) Find the magnitude of the acceleration of the system

(b) Find the magnitude of the force P that block m1 exerts on block m2

Fig 5.26 See Exercise (11)

F

1

m

P P a

2

m

(12) Repeat exercise 11 if block m2 is before block m1and the same force acts

on m2

(13) Two toys of masses m1 = 40 g and m2= 120 g are connected by a massless rope and lie on a horizontal frictionless surface as shown in Fig.5.27 The toys are

pulled to the right by a horizontal force of magnitude F = 0.04 N (a) Find the acceleration of the system (b) Find the magnitude of the tension force T in the

connecting rope

Fig 5.27 See Exercise (13)

2

a

(14) When the surface of exercise 13 is rough, it is found that the toys move with a constant velocity Find the common coefficient of kinetic frictionμkbetween the toys and the surface

(15) In Fig.5.28, the pulley is assumed massless and frictionless and rotates freely

about its axle The block has a mass m= 6 kg and the pulley is pulled to the right

by a horizontal force of magnitude F = 24 N As the block moves a distance s B

in time t, the pulley moves half that distance in the same time t, i.e., sP = s B /2 (a) Find the acceleration ratio aP /s B (b) Find the magnitudes of the tension

in the cord and the acceleration of the block if there is no friction between the

block and the surface (c) Answer part (b) assuming that the kinetic friction between the block and the surface isμk= 0.1.

Fig 5.28 See Exercise (15)

m

B

a

F

(16) A block of mass m1= 6 kg is hanging by a massless cord connected to another

block of mass m2 = 4 kg, which is also hanging by a massless cord, as shown

in Fig.5.29 (a) What is the tension in the cords when the system is at rest? (b) What is the tension in the cords when the two blocks are pulled up by the upper cord with an acceleration of 2 m/s2?

2

m

1

T

1

m

(17) In Fig.5.30, the pulley is assumed massless and frictionless and rotates freely

about its axle The blocks have masses m1 = 40 g and m2= 20 g, and block m1

is pulled to the right by a horizontal force of magnitude F = 0.03 N Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless

Fig 5.30 See Exercise (17)

a m

2 2

T

2

T

1

T

1

m

1

a

F