Q#27
A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is µ. The table does not move on the floor . Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table?
Answer:
Since the block moves on the table, the limiting friction has been achieved and the friction force on the block is µmg. Equal and opposite friction force will be applied by the block on the table = µmg, Since the table does not move on the floor, the limiting friction between the legs of the table and the floor has not been achieved, let this friction force be F. F will resist the force applied by the block on the table hence both will have opposite directions. Since the table is in equilibrium, sum of horizontal forces = zero,
i.e. F – µmg=0
F = µmg
We do not need the friction coefficient between the table and the floor or the mass of the table as is obvious from the above solution.
Q#28
Answer:
Let the acceleration of block M be 'a' and the tension in the string be T. The smaller block will have acceleration 'a' towards right and acceleration '2a' downwards. The Normal force on the block m = N = ma.
Friction force on block m by the block M = µ1N = µ1ma, in the vertical upward direction. Considering the forces in the vertical direction on block m.
mg – T – µ1ma = m(2a)
T = mg – µ1ma – 2ma
Now consider the vertical forces on the larger block M,
Downward forces are weight Mg and friction by smaller block m = µ1ma and Tension T at the pulley. So total downward force = Mg + µ1ma + T = Normal force by the ground N'
So friction force = µ2(Mg + µ1ma + T) its direction will be opposite to 'a'. Now consider the horizontal forces on the larger block M
2T – N – µ2(Mg + µ1ma + T) = Ma
2mg – µ1(2ma) – 4ma – ma – µ2(Mg + µ1ma + mg – µ1ma – 2ma) = Ma (putting the value of T)
2mg – µ2Mg – µ2mg+ µ2(2m)a = Ma + 5ma + µ1(2m)a
[2m – µ2(M + m)]g = Ma + 5ma + µ1(2m)a – µ2(2m)a
a = [2m – µ2(M + m)]g/[M + m{5 + 2(µ1 – µ2)}]
Q#29
A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force exerted by the wall on the block.
Answer:
Weight of the block = 2g = 20 N.
Normal force on the block by the wall when pushed by 40 N force = 40 N,
So, the weight of the block is just balanced by the friction force and block does not move if only the 40 N force is acting. When a horizontal force of 15 N parallel to the wall is applied, resultant of forces trying to move the block parallel to the wall will be resultant of weight 20 N and horizontal force 15 N both at right angles
= √(20² + 15²) = 25 N
which is more than the limiting static friction force. So, the block will move and move in the direction of this resultant force. Let x be the angle between resultant and 15 N force.
tan x = 20/15 = 4/3
x = 53° from the 15 N force.
Q#30
A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (figure 6-E11). Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s².
Answer:
(a)The man pushes both walls, in turn the walls also push him in equal and opposite directions. Considering, the horizontal forces in horizontal direction on the man. Only the normal forces by the walls on the man will act. Let normal force by walls be R and R'. Then R = R' at equilibrium. Hence the man pushes both walls with equal forces and the friction forces on both walls are also equal say F.
(b) Weight = 40g = 400 N.
In vertical direction 2F = 400
F = 200 N
Also, F = µR
200 = 0.8 R
R = 250 N
Q#31
Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is µ and that between the blocks is µ/2. Find the time elapsed before the smaller blocks separates from the bigger block.
Answer:
Since the friction between the blocks is half than the friction between road and the larger block, the smaller block will slide over the larger. Normal force on the smaller block = weight of the block = mg.
Friction force = ½ µmg
Retardation of the block = Friction force/mass = µg/2
See diagram below,
T = mg – µ1ma – 2ma
Now consider the vertical forces on the larger block M,
Downward forces are weight Mg and friction by smaller block m = µ1ma and Tension T at the pulley. So total downward force = Mg + µ1ma + T = Normal force by the ground N'
So friction force = µ2(Mg + µ1ma + T) its direction will be opposite to 'a'. Now consider the horizontal forces on the larger block M
2T – N – µ2(Mg + µ1ma + T) = Ma
2mg – µ1(2ma) – 4ma – ma – µ2(Mg + µ1ma + mg – µ1ma – 2ma) = Ma (putting the value of T)
2mg – µ2Mg – µ2mg+ µ2(2m)a = Ma + 5ma + µ1(2m)a
[2m – µ2(M + m)]g = Ma + 5ma + µ1(2m)a – µ2(2m)a
a = [2m – µ2(M + m)]g/[M + m{5 + 2(µ1 – µ2)}]
Q#29
A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force exerted by the wall on the block.
Answer:
Weight of the block = 2g = 20 N.
Normal force on the block by the wall when pushed by 40 N force = 40 N,
Limiting static Friction force on the block by the wall = µN = 0.5 x 40 N = 20 N.
So, the weight of the block is just balanced by the friction force and block does not move if only the 40 N force is acting. When a horizontal force of 15 N parallel to the wall is applied, resultant of forces trying to move the block parallel to the wall will be resultant of weight 20 N and horizontal force 15 N both at right angles
= √(20² + 15²) = 25 N
which is more than the limiting static friction force. So, the block will move and move in the direction of this resultant force. Let x be the angle between resultant and 15 N force.
tan x = 20/15 = 4/3
x = 53° from the 15 N force.
Q#30
A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (figure 6-E11). Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s².
Answer:
(a)The man pushes both walls, in turn the walls also push him in equal and opposite directions. Considering, the horizontal forces in horizontal direction on the man. Only the normal forces by the walls on the man will act. Let normal force by walls be R and R'. Then R = R' at equilibrium. Hence the man pushes both walls with equal forces and the friction forces on both walls are also equal say F.
(b) Weight = 40g = 400 N.
In vertical direction 2F = 400
F = 200 N
Also, F = µR
200 = 0.8 R
R = 250 N
Q#31
Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is µ and that between the blocks is µ/2. Find the time elapsed before the smaller blocks separates from the bigger block.
Answer:
Since the friction between the blocks is half than the friction between road and the larger block, the smaller block will slide over the larger. Normal force on the smaller block = weight of the block = mg.
Friction force = ½ µmg
Retardation of the block = Friction force/mass = µg/2
See diagram below,
Consider the forces on the larger block. Weights mg and Mg push the road, in turn the road applies a normal force equal and opposite in direction to the block.
So the normal force = (m + M)g
Since the system has been given a velocity and no other driving force acts, hence the friction force between larger block and the road will act against the motion to stop it.
This friction force = µ(m + M)g
Another friction force will act on the top surface applied by smaller block in the direction of motion = µmg/2
Total force against the motion = µ(m + M)g – µmg/2 = µmg/2 + µMg = µg(½m + M)
Retardation of larger block = µg(m/2 + M)/M
Difference in retardation of both blocks 'a'
= µg(m/2 + M)/M – ½µg
= µg{(½m + M)/M-1/2}
= µg{( m + 2M)/2M – ½}
= ½µg{(m + 2M)/M – 1}
= ½µg{(m + M)/M}
= Relative acceleration of the smaller block with respect to larger block.
As soon as the system is left after giving a velocity towards right, the smaller block starts to slip over the larger block. Here we consider the relative velocities of the smaller block w.r.t. larger. So initial velocity of smaller block, u = 0, distance to be traveled = l, acceleration a = µg(m + M)/2M, to get the time taken to travel this distance, t, we use the relation s = ut + ½at².
Here, l = 0 + ½[µg(m + M)/2M]t²
t²µg(m + M) = 4lM
t² = 4lM/µg(m + M)
t = √{4lM/µg(m + M)}
So, after this time the smaller block will separate from larger block.
So the normal force = (m + M)g
Since the system has been given a velocity and no other driving force acts, hence the friction force between larger block and the road will act against the motion to stop it.
This friction force = µ(m + M)g
Another friction force will act on the top surface applied by smaller block in the direction of motion = µmg/2
Total force against the motion = µ(m + M)g – µmg/2 = µmg/2 + µMg = µg(½m + M)
Retardation of larger block = µg(m/2 + M)/M
Difference in retardation of both blocks 'a'
= µg(m/2 + M)/M – ½µg
= µg{(½m + M)/M-1/2}
= µg{( m + 2M)/2M – ½}
= ½µg{(m + 2M)/M – 1}
= ½µg{(m + M)/M}
= Relative acceleration of the smaller block with respect to larger block.
As soon as the system is left after giving a velocity towards right, the smaller block starts to slip over the larger block. Here we consider the relative velocities of the smaller block w.r.t. larger. So initial velocity of smaller block, u = 0, distance to be traveled = l, acceleration a = µg(m + M)/2M, to get the time taken to travel this distance, t, we use the relation s = ut + ½at².
Here, l = 0 + ½[µg(m + M)/2M]t²
t²µg(m + M) = 4lM
t² = 4lM/µg(m + M)
t = √{4lM/µg(m + M)}
So, after this time the smaller block will separate from larger block.
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