Q#7
Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.Answer:
Let a push of F Newton is applied horizontally to move the block up the incline. Now a component of Fy = Fsin30° will also be added to the weight component mgcos30° pushing the block perpendicular to the plane, See the diagram below:
So, now the Normal force R= mgcos30°+Fsin30° and the Limiting frictional force
f =µR = µ(mgcos30°+ Fsin30°)
Total force acting down the incline on the block
= Limiting Frictional force + weight component down the incline
= µ(mgcos30° + Fsin30°) + mgsin30° (i)
To move the block up the incline the component of push along the incline (Fcos30°) should be equal to this force (i)
Equating the two we get
Fcos30°= µ(mgcos30° + Fsin30°) + mgsin30°
Solve this equation for F, which only is unknown in it,
Fcos30° = µmgcos30° + µFsin30° + mgsin30°
F(cos30° – µsin30°) = µmgcos30° + mgsin30°
F( ½√3 – 0.20 x ½) = 0.20 x 2 x 9.8 x ½√3 + 2 x 9.8 x ½
0.766F = 13.195
F = 17.23 N
Q#8
In a children park an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 10 m/s².
Answer:
Let mass of boy be M.
Normal force on the boy = Component of weight perpendicular to the plane = Mgcos45°
Frictional force on the boy = µMgcos45°
Component of weight along the plane = Mgsin45°
Net force on the boy down the incline = Mgsin45° – µMgcos45°
= Mg(1/√2 – 0.6/√2) = (0.4/√2)Mg
Acceleration of the boy = Force/mass
= (0.4/√2)g = 0.4 x 10/√2 = 2.82 m/s²
Q#9
A body starts slipping down an incline and moves half meter in half second. How long will it take it to move the next half meter?
Answer:
This problem has nothing to do with friction. From the given data we need to calculate the acceleration of the body and then calculate the unknown. Here we have,
Initial velocity u= 0 , Time taken = 0.5 s, Distance traveled s = 0.5 m,
To find acceleration 'a' we use the relation s = ut + ½at²
0.5 = 0 + ½a(0.5)²
0.5 = 0.5a(0.5)²
a = 1/(0.5)² = 4 m/s²
Let us calculate the time taken by the body to travel 1 m from the beginning, Now s = 1 m, u = 0, a = 4 m/s², t = ?
1 = 0 + ½(4)t² = 2t²
t² = 1/2 = 0.71 s
So to travel 1 m distance time taken is 0.71 s while given that time taken to travel first half meter = 0.5 s; Hence time taken to travel next half meter = 0.71 – 0.5 s = 0.21 s
Q#10
The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that if λ be the angle of friction and µ the coefficient of static friction, λ ≤ tan-1µ .
Answer:
See the diagram below:
F = Resultant contact force and λ = angle of friction,
Normal force R = Fcosλ and frictional force f = Fsinλ
But the limiting frictional force = µR = µFcosλ , which will be always greater than or equal to frictional force f = Fsinλ,
i.e. f ≤ µR
Fsinλ ≤ µFcosλ
tanλ≤µ
λ ≤ tan-1µ
Q#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.
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
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
= 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 x 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.
f =µR = µ(mgcos30°+ Fsin30°)
Total force acting down the incline on the block
= Limiting Frictional force + weight component down the incline
= µ(mgcos30° + Fsin30°) + mgsin30° (i)
To move the block up the incline the component of push along the incline (Fcos30°) should be equal to this force (i)
Equating the two we get
Fcos30°= µ(mgcos30° + Fsin30°) + mgsin30°
Solve this equation for F, which only is unknown in it,
Fcos30° = µmgcos30° + µFsin30° + mgsin30°
F(cos30° – µsin30°) = µmgcos30° + mgsin30°
F( ½√3 – 0.20 x ½) = 0.20 x 2 x 9.8 x ½√3 + 2 x 9.8 x ½
0.766F = 13.195
F = 17.23 N
Q#8
In a children park an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 10 m/s².
Answer:
Let mass of boy be M.
Normal force on the boy = Component of weight perpendicular to the plane = Mgcos45°
Frictional force on the boy = µMgcos45°
Component of weight along the plane = Mgsin45°
Net force on the boy down the incline = Mgsin45° – µMgcos45°
= Mg(1/√2 – 0.6/√2) = (0.4/√2)Mg
Acceleration of the boy = Force/mass
= (0.4/√2)g = 0.4 x 10/√2 = 2.82 m/s²
Q#9
A body starts slipping down an incline and moves half meter in half second. How long will it take it to move the next half meter?
Answer:
This problem has nothing to do with friction. From the given data we need to calculate the acceleration of the body and then calculate the unknown. Here we have,
Initial velocity u= 0 , Time taken = 0.5 s, Distance traveled s = 0.5 m,
To find acceleration 'a' we use the relation s = ut + ½at²
0.5 = 0 + ½a(0.5)²
0.5 = 0.5a(0.5)²
a = 1/(0.5)² = 4 m/s²
Let us calculate the time taken by the body to travel 1 m from the beginning, Now s = 1 m, u = 0, a = 4 m/s², t = ?
1 = 0 + ½(4)t² = 2t²
t² = 1/2 = 0.71 s
So to travel 1 m distance time taken is 0.71 s while given that time taken to travel first half meter = 0.5 s; Hence time taken to travel next half meter = 0.71 – 0.5 s = 0.21 s
Q#10
The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that if λ be the angle of friction and µ the coefficient of static friction, λ ≤ tan-1µ .
Answer:
See the diagram below:
 |
| Diagram for Answer - 10 |
Normal force R = Fcosλ and frictional force f = Fsinλ
But the limiting frictional force = µR = µFcosλ , which will be always greater than or equal to frictional force f = Fsinλ,
i.e. f ≤ µR
Fsinλ ≤ µFcosλ
tanλ≤µ
λ ≤ tan-1µ
Q#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.
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
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
= 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 x 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.
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