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EXERCISES (11-20)
11. Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.
Figure for problem 11
Answer: (a) Let the accelerations of 1.0 kg blocks be 'a'. So the acceleration of the 0.5 kg block will also be 'a'. If T be tension in the string attached to 0.5 kg block then force-acceleration equation for it is,
F=ma
→ 0.5g-T=0.5 a
→ T=0.5g-0.5a ..........................................(i)
Let tension in the string connecting 1.0 kg blocks be T'. Net force on the right hand side block = T-T'-µ*1.0g. Force acceleration equation for this block is
T-T'-µ*1.0g =1.0a
→ T-T'=0.2g+a ........................................(ii)
Similarly for left hand side block,
T'-0.2*1.0g=a
→ T'=0.2g+a ..........................................(iii)
Now we have three equations and three unknowns, T,T' and a.
Put the value of T and T' from (i) and (iii) in (ii).and we get,
0.5g-0.5a- 0.2g-a= 0.2g+a
→ 0.5a+a+a=0.5g-0.2g-0.2g
→ 2.5a=0.1g
→ a=0.04g m/s² =0.4 m/s² (Taking g=10 m/s²)
(b) T'=0.2g+a {from (iii)}
→ T'=0.2g+0.04g =0.24g N =2.4 N (Taking g=10 m/s²)
So the tension in the string connecting the 1.0 kg blocks = 2.4 N
(c) T=0.5g-0.5a {from (i)}
→ T=0.5g-0.5*0.04g =0.5g-0.02g
→ T=0.48g = 4.8 N (Taking g=10 m/s²)
So the tension in the string attached to 0.50 kg is 4.8 N.
12. If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 m/s², Find the friction coefficients at the two contacts with the blocks.
Figure for problem number 12
Answer: Friction force on the 2 kg block is µN =µ1*2g and it will act opposite to tension force of 16 N applied by the string. Since acceleration is 0.5 m/s², the force-acceleration equation is
F=ma
→16-µ1*2g=2*0.5
→ µ1*2g = 16-1=15
→ µ1 = 15/20 =0.75 (Taking g=10 m/s²)
So friction coefficient at 2 kg block is µ1 =0.75
Let us consider the 4 kg block, Normal force on the block
=mg cosθ =4g cos30° =4*10*√3/2 (Taking g=10 m/s²)
=20√3 N
Friction force on the block = µ2*20√3 N
Net force on the block along the plane =4g sinθ -16- µ2*20√3 N
From the force-acceleration equation,
4g sinθ -16 - µ2*20√3 =4*0.5
→ 40*sin30° - µ2*20√3 =16+2=18
→ 20 -18 = µ2*20√3
→ µ2 = 2/20√3 = 1/17.3 = 0.0578 ≈ 0.06
So friction coefficient at 4 kg block is µ2 = 0.06
13. The friction coefficient between the table and the block shown in figure (6-E4) is 0.2, find the tensions in the two strings.
Figure for problem number 13
Answer: Let the tension in the string connected to 15 kg block be T and to the 5 kg block be T'. If accelerations of block and weights be 'a' then applying force-acceleration equation (Derived from Newton's second Law of Motions) we get,
15g-T=15a →T=15g-15a ------------(i)
and T'-5g=5a →T'=5g+5a ------------(ii)
for block,
T-T'-µ*5g=5a
→ T-T'=0.2*5g+5a
→ 15g-15a-5g-5a = g+5a
{Putting value for T and T' from(i) and (ii)}
→ 25a = 9g
→ a = 90/25 =18/5 m/s² (Taking g=10 m/s²)
From (i),
T=15*10-15*18/5 =150-54 = 96 N
So tension in the left string =96 N
From (ii),
T'=5*10+5*18/5 =50+18 =68 N
So tension in the right string =68 N
14. The friction coefficient between a road and the Tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.
Answer: Let us first find the retardation of the vehicle from the given data. Initial velocity u=36 km/hr = 36000/3600 m/s =10 m/s, Final velocity v= 0, Distance traveled s = 5 m, if retardation is 'a',
v²=u²-2as
0 = 100-2a*5
→ a = 10 m/s²
Let the angle of inclination be θ and the mass of the vehicle be m. Normal force on the vehicle = mg.cosθ
Friction force =µN =4/3 * mg.cosθ
Net force along the incline against the motion
= 4/3 * mg.cosθ -mg.sinθ
Using force-acceleration relationship we get,
4/3 * mg.cosθ -mg.sinθ =ma = m*10 = 10m
→4g.cosθ - 3g.sinθ = 30
→40 cosθ -30 sinθ =30
→3 sinθ -4 cosθ +3 = 0
→3(1+sinθ) = 4 cosθ
→(1+sinθ)/cosθ =4/3
→(cos θ/2 + sinθ/2)/(cos θ/2 - sinθ/2)=4/3
→(1+tan θ/2)/(1-tan θ/2)=4/3
→3+3 tan θ/2 = 4-4 tan θ/2
→7 tan θ/2 = 1
→ tan θ/2 = 1/7 = tan 8.13°
→θ = 16°
So the maximum incline will be 16°.
15. The friction coefficient between an athlete's shoes and the ground is 0.90. Suppose a superman wears these shoes and and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop.
Answer: (a) Given µ= 0.90, if mass of superman = m, then maximum force of friction between shoes and ground = µmg. He can not apply more force on the ground to increase the speed because then the shoes will slip. If the acceleration of the superman be 'a', then,
a=force/mass = µmg/m = µg = 0.90x10 m/s² = 9 m/s².
Now initial velocity u = 0, Distance s = 50 m, a = 9 m/s², time t =?
from, s=ut+½at²
→ 50 = 0 + ½*9*t²
→ t² = 100/9
→ t =10/3 s = 3.3 s
So the minimum time he will have to take in completing the 50 m starting from rest = 3.3 s
(b) To stop he will have to depend on maximum friction force that will be again µmg and the retardation = µmg/m =µg =0.9x10 = 9 m/s². Speed of superman at completing 50 m = u+at =0+9*10/3 =30 m/s. Now this is the initial speed during the stopping, so u=30m/s, v=0 m/s, a= -9m/s²
from v=u+at
→ 0=30-9t
→ t=30/9 =10/3 s =3.3 s
So he will take exactly the same time to stop = 3.3 s.
16. A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30° (figure 6-E5). The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g=10 m/s².
Figure for problem number 16
Answer: Let mass of the car be m, Normal force on the car by the incline
=mgcos30° =mg√3/2
Maximum friction force =µ mg√3/2 =1/2√3*mg√3/2 =mg/4
Component of weight along slope = mg.sin30° = mg/2
Net force down the incline =mg/2-mg/4 =mg/4
Acceleration = Force/mass = mg/4m=g/4=10/4=2.5 m/s²
Now Initial velocity u=21.6 km/hr =21.6*1000/3600 m/s =6 m/s, Distance s =12.8 m, a= 2.5 m/s², final velocity v=?
from v²=u²+2as
v²=6²+2*2.5*12.8 =36+64=100
→ v=√100 = 10 m/s =10*3600/1000=36 km/hr
It is the minimum speed that the car can have at the bottom of the slope.
So no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr.
17. A car starts from rest on a half km long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one can not drive through the bridge in less than 10 s.
Answer: Let mass of the car = m, weight =mg=Normal force
Maximum force of friction available that will help drive the car =µmg = mg (Given µ=1)
So the maximum acceleration the car can have = Force/mass =mg/m = g =10 m/s²
Now initial velocity u = 0, distance s= 500 m, acceleration a=10 m/s², time taken t=? From s=ut+½at²
500=0+½*10*t²
→ t²=100
→ t = 10 s
So one can not drive through the bridge in less than 10 s.
18. Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is µ1, and that between the block of mass 4.0 kg and the incline is µ2. Calculate the acceleration of 2.0 kg block if (a) µ1 = 0.20 and µ2= 0.30 , (b) µ1 = 0.30 and µ2=0.20. Take g = 10 m/s².
Figure for problem number 18
Answer: Let us first calculate the accelerations of blocks assuming that they are not in contact. For 2 kg block,
Force down the incline = component of weight-friction force
=2g.sin30°-µ12g.cos30°
acceleration a1 =Force/mass = g.sin30°-µ1g.cos30° =10(½-µ1√3/2)
Similarly For 4 kg block,
acceleration a2 = g.sin30°-µ2g.cos30° = 10(½ -µ2√3/2)
(a) For µ1 = 0.20 and µ2= 0.30
a1 =10(½-µ1√3/2) =5-10*0.20*√3/2 =5-√3 =5-1.73 =3.27 m/s²
a2 =10(½-µ2√3/2) =5-10*0.30*√3/2 =5-1.5√3 =2.40 m/s²
Since a1> a2 , So sliding together upper and lower blocks will push each other, let the joint acceleration be 'a'. Now considering the 2 kg block an extra force 'F' exerted by the lower block will act on it. Now force on the block =2g.sin30°-µ12g.cos30°-F
and acceleration a= (2g.sin30°-µ12g.cos30°-F)/2 ...............(i)
Considering 4 kg block,
force on the block =4g.sin30°-µ2 4g.cos30°+F
and acceleration a= (4g.sin30°-µ2 4g.cos30°+F)/4 ..............(ii)
Equating (i) and (ii)
(2g.sin30°-µ12g.cos30°-F)/2 = (4g.sin30°-µ2 4g.cos30°+F)/4
→ sin30°-µ1.cos30°-F/2g = sin30°-µ2 cos30°+F/4g
→ F/4g+F/2g = µ2 cos30° -µ1.cos30°
→ 3F/4g = (0.30-0.20)√3/2 =0.1*√3/2
→ F = 4g/3* 0.1*√3/2 =40*0.1√3/6 =4√3/6 =2√3/3
So acceleration a= (2g.sin30°-µ12g.cos30°-F)/2 from..(i)
=g.½-0.20g√3/2-√3/3
=5-√3-√3/3
=5-4√3/3=2.70 m/s²
Acceleration of of 2.0 kg block a =2.70 m/s².
(b) For µ1 = 0.30 and µ2= 0.20
a1 =10(½-µ1√3/2) =5-10*0.30*√3/2 =2.41 m/s²
a2 =10(½-µ2√3/2) =5-10*0.20*√3/2 =3.27 m/s²
=g.½-0.20g√3/2-√3/3
=5-√3-√3/3
=5-4√3/3=2.70 m/s²
Acceleration of of 2.0 kg block a =2.70 m/s².
(b) For µ1 = 0.30 and µ2= 0.20
a1 =10(½-µ1√3/2) =5-10*0.30*√3/2 =2.41 m/s²
a2 =10(½-µ2√3/2) =5-10*0.20*√3/2 =3.27 m/s²
Since acceleration of lower block is more than the upper one, so both blocks would not remain in contact. Acceleration of of 2.0 kg block a1 =2.41 m/s².
19. Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is µ. Find the acceleration of the system and the force by the rod on one of the blocks.
Answer: Let the force by the rod on one of the blocks be F and acceleration of the system =a.
For upper block of mass M1 , Net Force down the plane
= M1 g.sinθ-µ M1 g.cosθ-F
Acceleration a =(M1 g.sinθ-µ M1 g.cosθ-F)/M1
=g.sinθ-µg.cosθ-F/M1
For lower block M2, F will act along the motion, So net force down the plane =Weight component-Friction force+F
= M2 g.sinθ-µ M2 g.cosθ+F
For lower block M2, F will act along the motion, So net force down the plane =Weight component-Friction force+F
= M2 g.sinθ-µ M2 g.cosθ+F
Acceleration a =(M2 g.sinθ-µ M2 g.cosθ+F)/M2
=g.sinθ-µg.cosθ+F/M2
Since both blocks are connected, their accelerations will be the same. Equating the accelerations,
g.sinθ-µg.cosθ-F/M1 =g.sinθ-µg.cosθ+F/M2
→ F/M1+F/M2 =0
→ F(1/M1+1/M2) =0
→ F = 0
So the force by the rod on one of the blocks is zero.
Now acceleration a = g.sinθ-µg.cosθ+F/M1
= g.(sinθ-µcosθ)
Since both blocks are connected, their accelerations will be the same. Equating the accelerations,
g.sinθ-µg.cosθ-F/M1 =g.sinθ-µg.cosθ+F/M2
→ F/M1+F/M2 =0
→ F(1/M1+1/M2) =0
→ F = 0
So the force by the rod on one of the blocks is zero.
Now acceleration a = g.sinθ-µg.cosθ+F/M1
= g.(sinθ-µcosθ)
20. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is µ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?
Answer: Let the pull force F be applied to the block at an angle θ from the horizontal, See figure below:--
Diagram for Problem 20
We resolve force F in vertical and horizontal directions as Fsinθ and Fcosθ. Force applied by the block on the surface
= Weight- vertical component of F
= Mg-F.sinθ = Normal force by the surface on the block
Given Static friction coefficient =µ,
So Maximum Friction force =µ(Mg-F.sinθ) {In horizontal direction}
Hence to move the block horizontal component of F must be just greater than this friction force,
i.e. F.cosθ=µ(Mg-F.sinθ)
→ F.cosθ+µF.sinθ=µMg
→ F= µMg/(cosθ+µsinθ)
Since µ,M and g are constant in this case, so numerator of this expression is fixed. The force F will be minimum if denominator of the above expression is maximum,
i.e. (cosθ+µsinθ) should be maximum. To find the θ for which it is maximum we differentiate it with respect to θ and equate to zero.
Let y= (cosθ+µsinθ)
dy/dθ=-sinθ+µ.cosθ=0
→ tanθ=µ
→ θ=tan-1µ
Since dy/dθ is zero for both at maxima and minima, to know whether it is maxima and minima we again differentiate it,
d²y/dθ²= -(cosθ+µ.sinθ)
For a positive value of µ, 0°<θ<90°, and cosθ+µsinθ will be positive, Hence d²y/dθ² = -ve which gives that it is a maxima value.
So the force should be applied at an angle θ=tan-1µ from the horizontal to move the block.
And minimum force needed to slide the block is
= µMg/(cosθ+µsinθ)
= µMg.secθ/(1+µtanθ) {Divide numerator &denominator by cosθ}
= µMg√sec²θ/(1+µ.µ)
= µMg√(1+tan²θ)/(1+µ.µ)
= µMg√(1+µ²)/(1+µ²)
= µMg/√(1+µ²)
===<<O>>===
HC Verma's Concepts of Physics, Chapter-7, Circular Motion,
Click here for → Questions for short Answer
Click here for → Objective I
Click here for → Objective II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
HC Verma's Concepts of Physics, Chapter-6, Friction,
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
-------------------------------------------------- ----------------------------- Click here for → Friction OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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