Solved Examples on Laws of Motion

Question 1:A chain consisting of five links, each with mass 100 g, is lifted vertically with a constant acceleration of 2.50 m/s^{2}, as shown in below figure. Find,

(a)The forces acting between adjacent links.

(b)The forceFexerted on the top link by the agent lifting the chain.

(c)The net force on each link.

Concept:Force acting on the object (

F) is equal to the product of mas of the (m) and acceleration of the auto mobile (a).So,

F_{net}=maWeight of the automobile (

W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g= 9.81 m/s^{2}).

W=mg

Solution:

(a)The net force on each link is the same. To obtain the net forceF_{net}on each link, substitute 0.100 kg formand 2.50 m/s^{2}forain the equationF_{net}=ma,

F_{net}=ma=(0.100 kg)(2.50 m/s

^{2})=(0.250 kg.m/s

^{2}) (1 N/1 kg.m/s^{2})=0.250 N

To obtain the weight

Wof the each link, substitute 0.100 kg formand 9.81 m/s^{2}forgin the equationW=mg,

W=mg=(0.100 kg)( 9.81 m/s

^{2})= (0.981 kg.m/s

^{2}) (1 N/1 kg.m/s^{2})=0.981 N

On each link (except the top or bottom link) there is a weight

W, an upward forceUfrom the link above, and a downward forceDfrom the link below.So,

F_{net}=U-D-WThen,

U=F_{net}+W+D=(0.250 N)+(0.981 N)+

D=1.23 N+

DFor the bottom link,

D=0 .So,

U=1.23 N+D=1.23 N+0

=1.23 N

For the link above,

U= 1.23 N+1.23 N= 2.46 N

For the next link above,

U= 1.23 N+2.46 N= 3.69 N

For the next link above,

U= 1.23 N+3.69 N= 4.92 N

From the above observation we conclude that, the force

Fexerted on the top link by the agent lifting the chain would be 4.92 N.

(b)To obtain the upward forceUfor the top link, substitute 4.92 N forDin the equationU=1.23 N+D,

U=1.23 N+D=1.23 N+4.92 N

=6.15 N

From the above observation we conclude that, the upward force

Ufor the top link would be 6.15 N.

(c)The net force on each link is the same. To obtain the net forcesF_{net}acting between adjacent on each link, substitute 0.100 kg formand 2.50 m/s^{2}forain the equationF_{net}=ma,

F_{net}=ma=(0.100 kg)(2.50 m/s

^{2})=(0.250 kg.m/s

^{2}) (1 N/1 kg.m/s^{2})=0.250 N

From the above observation we conclude that, the forces acting between adjacent links would be 0.250 N.

Question 2:(a) Two 10-lb weights are attached to a spring scale as shown in Fig. 1a. What is the reading of the scale?(b) A single 10-lb weight is attached to a spring scale which itself is attached to a wall, as shown in figure 1b. What is the reading of the scale? (Ignore the weight of the scale.)

Solution:a) Let

W_{1}is the weight of the left side block (W_{1}= 10-lb),a_{1}is the upward acceleration of the block andm_{1}is the mass of the block of the left hand side.The upward force due to the upward acceleration

a_{1}will be,

F_{1}=m_{1}a_{1}…… (1)If

Tis the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be,

m_{1}a_{1}=W_{1}-T…… (2)Let

W_{2}is the weight of the right side block (W_{2}= 10-lb),a_{2}is the downward acceleration of the block andm_{2}is the mass of the block of the right hand side.The downward force due to the downward acceleration

a_{2}on the right side block will be,

F_{2}=m_{2}a_{2}…… (3)If

Tis the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be,

m_{2}a_{2}=W_{2}-T…… (4)FBD:-

Subtracting equation (4) from equation (2),

m_{1}a_{1}-m_{2}a_{2}=W_{1}-T–T+W_{2}…… (5)Since,

W_{1}=W_{2}=10-lb and there is no relative motion, the left side of the equation (5) will be zero.So,

W_{1}-T–T+W_{2}= 02

T= (W_{1}+W_{2})= (10-lb+10-lb)

So, T = 20-lb/2

=10-lb …… (6)

From equation (6) we observed that, the reading of the scale will be 10-lb.

b) When a single 10-lb weight (

W=10-lb) is attached to a spring scale which itself is attached to a wall, the net exerted force (F_{net}) on the block will be equal to weight of the block (W) minus tension exerted on the block (T).So,

F_{net}=W-T…… (7)Since the weight is attached to a spring scale which itself is attached to a wall, therefore the net force will be zero (

F_{net}=0).Putting the value of

F_{net}in equation (7), we get,

W-T=0

T=W= 10-lb …… (8)

From equation (8) we observed that, the reading of the scale will be 10-lb.

Question 3:-A 1200-kg car is being towed up an 18^{°}incline by means of a rope attached to the rear of a truck. The rope makes an angle of 27^{°}with the incline. What is the greatest distance that the car can be towed in the first 7.5 s starting from rest if the rope has a breaking strength of 4.6 kN? Ignore all the resistive forces on the car. See the low figure.

Concept:The figure below shows the car of mass

Mtowed by another car through a rope of breaking strengthT.

Assume that the

xaxis is along the length of the ramp andyaxis is perpendicular to it such that the tension in the rope along thexaxis is whereas the tension in the rope along theyaxis is .Apply Newton’s second law of motion along the

xaxis,

Tcos (27^{°}) –Mgsin (18°) =Ma_{x}Here

a_{x}is the acceleration of the car of massMalong thexaxis andgis the acceleration due to gravity.

a_{x}=Tcos (27^{°}) –Mgsin (18°)/MFrom this acceleration, one can calculate the distance travelled by the car in time

tusing relation,

x= ½a_{x}t^{2}

Solution:Substitute 1200 kg for

M, 9.8 m/s^{2}for g and 4.6 kN forTin equation,

a= [_{x}Tcos (27^{°}) –Mgsin (18^{°})]/M

a= [(4.6 kN) (10_{x}^{3}N/1 kN) cos (27^{°}) – (1200 kg) (9.81 m/s^{2})sin (18^{°})]/[1200 kg]= {(4098.6 N) [(1 kg.m/s

^{2})/1 N]/[1200 kg]}-3.02 m/s^{2}= 3.41 m/s

^{2}- 3.02 m/s^{2}= 0.38 m/s

^{2}Therefore, the acceleration of the car towed up by another car with a rope of breaking strength 4.6 kN is 0.38 m/s

^{2}.Substitute 0.38 m/s

^{2}fora_{x}and 7.5 s fortin equationx= ½a_{x}t^{2},

x= ½ (0.38 m/s^{2}) (7.5 s)^{2}= 10.8 m

Therefore, the distance by which the car can move in 7.5 s is 10.8 m.

Question 4:-Someone exerts a forceFdirectly up on the axle of the pulley shown in below figure. Consider the pulley and string to be mass less and the bearing frictionless. Two objects,m_{1}of mass 1.2 kg andm_{2}of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The objectm_{2}is in contact with the floor.(a)What is the largest value of force may have so thatm_{2}will remain at rest on the floor?(b)What is the tension in the string if the upward forceFis 110 N?(c)With the tension determined in part (b), what is the acceleration ofm_{1}?

Concept:The figure below shows the forces acting in the block-pulley system:

It is important to note that the tension pulling the pulley down is 2

Twhereas the force pushing it upwards isF. Therefore, the block with massm_{2}will stay on floor when

F≤ 2T, and the tension in the stringTis equal to the weight of the block on floor.The net force on block with mass

m_{1}is given as:

F_{net}=F-m_{1}gThus, the acceleration of block with mass

m_{1}is:

a= (T-m_{1}g)/m_{1}

Solution:

(a)Substitute 1.9 kg form_{2}, and 9.8 m/s^{2}forgin equationT=m_{2}g,

T= (1.9 kg) (9.8 m/s^{2})= (18.62 kg. m/s

^{2}) [1 N/(1kg. m/s^{2})]= 18.62 N

The maximum force

Ffor which the block stays on the floor isF= 2T.Substitute 18.62 N for

Tin equationF= 2T,

F= 2 (18.62 N)= 37.24 N

Round off to two significant figures

F= 37 NTherefore, the maximum force

Ffor which block with massm_{2}stays on floor is 37 N.

(b)The tension in the string is related to upward force as:

T=F/2Substitute 110 N for

F,

T=F/2= 110 N/2

= 55 N

Therefore, the tension in the string corresponding to upward force of 110 N is 55 N.

(c)Substitute 55 N forT, 1.2 kg form_{1}and 9.8 m/s^{2}forgin equationa=T-m_{1}g/m_{1}

a= [55 N- (1.2 kg) (9.8 m/s^{2})]/1.2 kg= {(55 N) [(1 kg.m/s

^{2})/1 N]- (11.76 kg.m/s^{2})}/1.2 kg= 36 m/s

^{2}Therefore, the acceleration of the block with mass

m_{1}is 36 m/s^{2}.

Question 5:-The coefficient of static friction between the tires of a car and a dry road is 0.62. The mass of the car is 1500 kg. What maximum braking force is obtainable

(a)on a level road and

(b)on an 8.6°downgrade?

Concept:Assume that the mass of the car is

mand the coefficient of friction between the tires of the car and the road isµ_{s}.The magnitude of maximum braking force obtainable on the level road is equal to the magnitude of the static frictional force between the tires and the road. The magnitude of the frictional force is,

f=µ_{s}NHere

Nis the magnitude of normal force on the car.

(a)The car is in vertical equilibrium on the level road therefore the normal force on the car should be equal to its weight (mg) in magnitude, and must act in opposite direction, that is

N=mgSubstitute

N=mgin equationf=µ_{s}N,

f=µ_{s}N=

µ_{s}mgTherefore, the magnitude of maximum obtainable braking force is .

(b)When the car is on the inclined road, the normal force is not equal to the weight of the car but to its componentmgcosθ, that is

N=mgcosθHere

θis the angle of inclination of the road.Substitute

N=mgcosθin equationf=µ_{s}N,

f=µ_{s}N=

µ_{s}mgcosθTherefore, the magnitude of maximum obtainable braking force on the inclined road is

µ_{s}mgcosθ.

Solution:

(a)Substitute 0.62 forµ_{s}, 1500 kg formand 9.81 m/s^{2}forgin equationF=µ_{s}mg,

F=µ_{s}mg= (0.62) (1500 kg) (9.81 m/s

^{2})= (9123.3 kg.m/s

^{2}) [1 N/(1 kg.m/s^{2})]= 9123. 3 N

Round off to four significant figures,

F= 9123 NTherefore, the maximum obtainable braking force on the level road is 9123 N.

(b)Substitute 0.62 forµ_{s}, 1500 kg form, 8.6^{°}forθand 9.81 m/s^{2}forgin equation

F=µ_{s}mgcosθ,

F=µ_{s}mgcosθ

= (0.62) (1500 kg) (9.81 m/s^{2}) cos (8.6^{°})= (9020 kg. m/s

^{2}) [1 N/(1 kg.m/s^{2})]= 9020 N

Therefore, the maximum obtainable braking force on the inclined road is 9020 N.

Question 6:-A railroad flat car of weightWcan roll without friction along a straight line track. Initially, a man of weightwis standing on the car, which is moving to the right with speedv_{0}. What is the change in velocity of the car if the man runs to the left (below figure) so that his speed relative to the car isv_{rel}just before he jumps off at the left end?

:

ConceptMomentum of the body

pis equal to the mass of the bodymtimes velocity of the bodyv.So,

p=mvIn accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system.

Consider the initial momentum of the man is

p_{i,m}, initial momentum of the car isp_{i,c}, final momentum of the man isp_{f,m}and final momentum of the car isp_{f,c}.Initial momentum of the man is equal to the initial momentum of the car.

So, initial momentum of the man will be,

p_{i,m}=m_{m}v_{i,c}and final momentum of the man

p_{f,m}will be,

p_{f,m}=m_{m}(v_{f,c}–v_{rel})Here, speed of the man relative to the car is

v_{rel}.Again,

p_{f,c}=m_{c}v_{f,c}

p_{i,c}=m_{c}v_{i,c}Here, mass of the man is

m_{m}, mass of the cart ism_{c}, initial velocity of the man isv_{i,m}and cart isv_{i,c}, and final velocity of the man isv_{f,m}and cart isv_{f,c}.

Solution:So applying conservation of momentum to this system, the sum of the initial momentum of the man and car will be equal to the sum of the final momentum of the man and cart.

p_{f,m}+p_{f,c}=p_{i,m}+p_{i,c}Substitute,

m_{m}(v_{f,c}–v_{rel}) forp_{f,m},m_{c}v_{f,c}forp_{f,c},m_{m}v_{i,c}forp_{i,m}andm_{c}v_{i,c}forp_{i,c}in the equationp_{f,m}+p_{f,c}=p_{i,m}+p_{i,c},

p_{f,m}+p_{f,c}=p_{i,m}+p_{i,c}

m_{m}(v_{f,c}–v_{rel})+m_{c}v_{f,c}=m_{m}v_{i,c}+m_{c}v_{i,c}(

m_{m}+m_{c})v_{f,c}-m_{m}v_{rel}= (m_{m}+m_{c})v_{i,c}(

m_{m}+m_{c})v_{f,c}- (m_{m}+m_{c})v_{i,c}=m_{m}v_{rel}(

m_{m}+m_{c}) (v_{f,c}-v_{i,c}) =m_{m}v_{rel}(

m_{m}+m_{c}) (Δv_{c}) =m_{m}v_{rel}Or, Δ

v_{c}=m_{m}v_{rel}/(m_{m}+m_{c})= [

w/g]v_{rel}/[(w/g) + (W/g)] (Since,m_{m}=w/gandm_{c}=W/g)=

wv_{rel}/(w+W)Here, weight of the man is

wand weight of the railroad flat car isW.From the above observation we conclude that, the change in velocity of the car if the man runs to the left would be

wv_{rel}/(w+W).

Question 7:-A man of massmclings to a rope ladder suspended below a balloon of massM; see below figure. The balloon is stationary with respect to ground.(a) If the man beings to climb the ladder at a speed

v(with respect to ladder), in what direction and with what speed (with respect to Earth) will the balloon move?(b) What is the state of motion after the man stops climbing?

Concept:When the man climbs up on a ladder mounted on a balloon, no external force acts on the system (the effect of gravity is neglected). Therefore, the center of mass of the balloon-man system does not accelerate.

Solution:

(a)Assume that the mass of the man ismclimbing on the ladder with velocityvand the mass of the balloon isMmoving downward with velocityv_{b}.From definition of center of mass, one can write

(

M+m)v_{cm}=Mv_{b}+m(v+v_{b})Substitute

v_{cm }= 0,0 =

Mv_{b}+m(v+v_{b})

v_{b}= -mv/M+mThe negative sign in the equation highlights the fact that the balloon moves in the direction opposite to that of man, which is downwards in this case.

It is important to that when the balloon moves downwards with velocity

v_{b}and the man moves upward with velocityv, the velocity with which the man appears to move relative to ground isv+v_{b}.Therefore, the velocity with which the balloon moves downward is given as

v_{b}= -mv/M+m.

(b)When the man stops climbing up the ladder, the balloon also stops moving downwards and eventually come to rest.