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How do you evaluate the tension force if the coefficient of friction between the objects K and L is $0.6$?
How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?
So the system is accelerating, whence we have to consider that
$$sum F_x = m_1a$$
$$F_k - T = 2a $$
$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$
For the object L,
$$sum F_x = m_2a$$
$$F - F_k = 6a $$
$$10 - 12 = 6a implies a = -dfrac{1}{3}$$
Plugging $a$ into the first equation
$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$
However, there won't be an integer solution from what I got above. Could you assist me?
physics
asked Nov 17 '18 at 15:14
Enzo
1216
add a comment |
How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?
So the system is accelerating, whence we have to consider that
$$sum F_x = m_1a$$
$$F_k - T = 2a $$
$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$
For the object L,
$$sum F_x = m_2a$$
$$F - F_k = 6a $$
$$10 - 12 = 6a implies a = -dfrac{1}{3}$$
Plugging $a$ into the first equation
$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$
However, there won't be an integer solution from what I got above. Could you assist me?
physics
asked Nov 17 '18 at 15:14
Enzo
1216
1
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23
add a comment |
How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?
So the system is accelerating, whence we have to consider that
$$sum F_x = m_1a$$
$$F_k - T = 2a $$
$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$
For the object L,
$$sum F_x = m_2a$$
$$F - F_k = 6a $$
$$10 - 12 = 6a implies a = -dfrac{1}{3}$$
Plugging $a$ into the first equation
$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$
However, there won't be an integer solution from what I got above. Could you assist me?
physics
asked Nov 17 '18 at 15:14
Enzo
1216
How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?
So the system is accelerating, whence we have to consider that
$$sum F_x = m_1a$$
$$F_k - T = 2a $$
$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$
For the object L,
$$sum F_x = m_2a$$
$$F - F_k = 6a $$
$$10 - 12 = 6a implies a = -dfrac{1}{3}$$
Plugging $a$ into the first equation
$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$
However, there won't be an integer solution from what I got above. Could you assist me?
physics
physics
asked Nov 17 '18 at 15:14
Enzo
1216
asked Nov 17 '18 at 15:14
Enzo
1216
asked Nov 17 '18 at 15:14
Enzo
1216
asked Nov 17 '18 at 15:14
Enzo
1216
asked Nov 17 '18 at 15:14
Enzo
1216
1216
1
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23
add a comment |
1
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23
1
1
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23
add a comment |
2 Answers
2
active
oldest
votes
To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.
Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.
Since the tension balances the frictional force, the tension is $10text{N}$.
answered Nov 17 '18 at 23:35
John Douma
5,36211319
add a comment |
It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.
1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.
2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.
Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.
Since the tension balances the frictional force, the tension is $10text{N}$.
answered Nov 17 '18 at 23:35
John Douma
5,36211319
add a comment |
To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.
Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.
Since the tension balances the frictional force, the tension is $10text{N}$.
answered Nov 17 '18 at 23:35
John Douma
5,36211319
add a comment |
To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.
Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.
Since the tension balances the frictional force, the tension is $10text{N}$.
answered Nov 17 '18 at 23:35
John Douma
5,36211319
To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.
Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.
Since the tension balances the frictional force, the tension is $10text{N}$.
answered Nov 17 '18 at 23:35
John Douma
5,36211319
answered Nov 17 '18 at 23:35
John Douma
5,36211319
answered Nov 17 '18 at 23:35
John Douma
5,36211319
answered Nov 17 '18 at 23:35
John Douma
5,36211319
5,36211319
add a comment |
add a comment |
It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.
1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.
2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
add a comment |
It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.
1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.
2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
add a comment |
It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.
1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.
2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.
1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.
2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
edited Nov 20 '18 at 20:06
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
answered Nov 17 '18 at 21:39
Rafa Budría
5,5651825
5,5651825
add a comment |
add a comment |
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1
Who says the solution must be integer?
– Sean Roberson
Nov 17 '18 at 15:19
@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 '18 at 15:23