A weightless string is suspend on a frictionless pulley, a mass of 6 kg. is
suspend at one end and mass of 10 kg. is suspended on the other end of the
string, the tension in the string will be :
a up on 6 kg mass, so down on 10 kg mass
6 kg mass
T - 6 g = 6 a
10 kg mass
10 g - T = 10 a
add
4 g = 16 a
a = g/4
then go back
T - 6 g = 6 g/4
T = 7.5 g = 7.5(9.81)
To find the tension in the string, we need to consider the forces acting on each mass. Here are the steps to solve the problem:
1. Identify the forces acting on each mass:
- For the 6 kg mass, there is only the force of gravity acting downward (mg, where g is the acceleration due to gravity).
- For the 10 kg mass, there is the force of gravity acting downward (mg) and the tension in the string acting upward.
2. Write the equations for the forces (using Newton's second law, F = ma) for each mass:
- For the 6 kg mass: mg = 6a, where a is the acceleration of the system.
- For the 10 kg mass: mg - T = 10a, where T is the tension in the string and a is the acceleration.
3. Since the pulley is frictionless, the two masses will have the same acceleration (a) and opposite directions. So we can set up an equation relating their accelerations:
6a = 10a
4. Solve the equation from step 3 for acceleration (a):
6a = 10a
6a - 10a = 0
-4a = 0
a = 0
5. Substitute the acceleration (a) into one of the equations from step 2 to find the tension (T) in the string:
T = mg - 10a
T = 10(9.8) - 10(0)
T = 98 N
Therefore, the tension in the string is 98 N.
To find the tension in the string, we need to consider the forces acting on each mass.
Let's label the mass of 6 kg as m1 and the mass of 10 kg as m2. In this case, m2 is greater than m1.
For m1 (6 kg):
- The weight of m1 is given by W1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The tension in the string pulling m1 upwards is T.
For m2 (10 kg):
- The weight of m2 is given by W2 = m2 * g.
- The tension in the string pulling m2 downwards is also T.
Since the string is weightless and frictionless, the tension is the same throughout the string.
Now, we can set up the equations of motion for each mass using Newton's second law (F = ma):
For m1:
T - W1 = m1 * a ... (Equation 1)
For m2:
W2 - T = m2 * a ... (Equation 2)
Since the pulley is assumed to be frictionless, we can assume that the accelerations of m1 and m2 are the same. Therefore, a1 = a2 = a.
Now, let's substitute the expressions for W1, W2, and rearrange the equations:
T - (m1 * g) = m1 * a ... (Equation 1)
(m2 * g) - T = m2 * a ... (Equation 2)
Now, we can solve these equations simultaneously to find the value of T.
Adding Equation 1 and Equation 2:
T - (m1 * g) + (m2 * g) - T = m1 * a + m2 * a
(m2 * g) - (m1 * g) = (m1 + m2) * a
g * (m2 - m1) = (m1 + m2) * a
Now, solving for a:
a = (g * (m2 - m1)) / (m1 + m2)
Plugging in the given values:
a = (9.8 m/s^2 * (10 kg - 6 kg)) / (6 kg + 10 kg)
a = (9.8 m/s^2 * 4 kg) / 16 kg
a = 2.45 m/s^2
Now, we can substitute the value of a back into either Equation 1 or Equation 2 to find the tension T:
T = m1 * a + W1
T = 6 kg * 2.45 m/s^2 + (6 kg * 9.8 m/s^2)
T = 14.7 N + 58.8 N
T = 73.5 N
Therefore, the tension in the string is 73.5 Newtons.