Three blocks, with masses m1 = 5.90 kg, m2 = 5.50 kg, and m3 = 7.10 kg, are pull on a horizontal frictionless surface by a 18.00 N force that makes a 39 o angle (θ) with the horizontal. What is the magnitude of the tension between the m2 and m3 blocks?

Well, it seems like we have a weighty situation here! Let's sort it out, shall we?

To find the magnitude of the tension between the m2 and m3 blocks, we first need to break down the forces acting on them.

Since the surface is frictionless, we only have the force pulling the blocks as the horizontal force. This force can be resolved into the x-component and the y-component.

The x-component of the force is given by Fx = F * cos(θ), where F is the force magnitude. Plugging in the values, we have Fx = 18.00 N * cos(39°).

Now, let's divide the force among the blocks based on their masses. The tension between m2 and m3 will be equal to the force acting on them in the x-direction.

Since the force is the same for both blocks, the magnitude of the tension between m2 and m3 will be the same as Fx.

So, the magnitude of the tension between the m2 and m3 blocks is approximately 18.00 N * cos(39°).

I hope the answer doesn't pull any strings or cause any weighing on your mind!

To find the magnitude of the tension between the m2 and m3 blocks, we need to analyze the forces acting on the blocks and apply Newton's second law of motion.

First, let's break down the given force into its horizontal and vertical components. The horizontal component (F_x) is given by F * cos(θ), and the vertical component (F_y) is given by F * sin(θ).

F_x = 18.00 N * cos(39 o)
F_y = 18.00 N * sin(39 o)

Next, let's examine the forces acting on each block individually:

For m1:
- The force of tension (T_1) pulling to the right.
- The force of weight (m1 * g) pulling downward.

For m2:
- The force of tension (T_2) pulling to the right.
- The force of weight (m2 * g) pulling downward.
- The force of tension (T_3) pulling to the left.

For m3:
- The force of tension (T_3) pulling to the right.
- The force of weight (m3 * g) pulling downward.

Since the surface is frictionless, there is no horizontal net force. Therefore, the magnitudes of T_1 and T_2 must be equal (T_1 = T_2). This is because the horizontal component of the tension in both m1 and m2 must be equal to counteract the applied force.

We can now write the equations for the forces acting on each block:

For m1 along the horizontal direction:
T_1 - T_3 = m1 * a_x (equation 1)

For m2 along the horizontal direction:
T_2 - T_3 = m2 * a_x (equation 2)

For m3 along the horizontal direction:
T_3 = m3 * a_x (equation 3)

Since the surface is frictionless, the acceleration (a_x) is the same for all three blocks.

Now, let's add the vertical forces for m1 and m2:

For m1 along the vertical direction:
N1 - m1 * g = 0 (equation 4)

For m2 along the vertical direction:
N2 - m2 * g = 0 (equation 5)

Where N1 and N2 are the normal forces exerted on m1 and m2, respectively. Since the surface is frictionless, the normal forces are equal to the weight of each block.

Now, let's solve the system of equations:

From equation 4, we get:
N1 = m1 * g

From equation 5, we get:
N2 = m2 * g

From equations 3, 4, and 5, we can solve for T_3:
T_3 = m3 * a_x = N1 = N2 = m1 * g = m2 * g

Next, let's substitute the values of m1, m2, and m3 into the equation for T_3:

T_3 = (7.10 kg) * (9.8 m/s^2) = 69.58 N

Finally, the magnitude of the tension between the m2 and m3 blocks is 69.58 N.

To find the magnitude of tension between the m2 and m3 blocks, we can follow these steps:

Step 1: Analyze the forces acting on the system:
- There is the applied force of 18.00 N, which makes an angle of 39° with the horizontal.
- The tension force, T, between the m2 and m3 blocks.
- The gravitational forces acting on each block, given by their masses (m1, m2, and m3) and the acceleration due to gravity (9.8 m/s²).

Step 2: Resolve the applied force into its horizontal and vertical components:
- The horizontal component, F_horizontal, is equal to F_applied * cos(θ).
- The vertical component, F_vertical, is equal to F_applied * sin(θ).

Step 3: Calculate the net force in the horizontal direction:
- The net horizontal force is given by F_horizontal - T because both forces act in opposite directions.

Step 4: Apply Newton's second law to find the acceleration of the system:
- The net horizontal force is equal to the mass of the system (m1 + m2 + m3) multiplied by the acceleration (a).

Step 5: Calculate the net force in the vertical direction:
- The net vertical force is given by the sum of the gravitational forces acting on each block, which is m1 * g + m2 * g + m3 * g.

Step 6: Apply Newton's second law in the vertical direction:
- The net vertical force is equal to the total mass of the system multiplied by the acceleration due to gravity (m1 + m2 + m3) * g.

Step 7: Solve the two equations simultaneously for the acceleration and tension force:
- Equate the net horizontal force (m1 + m2 + m3) * a to F_horizontal - T.
- Equate the net vertical force (m1 + m2 + m3) * g to F_vertical.

Step 8: Solve the equations to find the acceleration and tension force:
- Substitute the values and solve for a and T. Once you find the value of a, you can use it to find the tension force.

Following these steps, you can find the magnitude of the tension force between the m2 and m3 blocks.

frictionless?

pulling force=18*cos39

Netforce=total mass*a
a=18cos39/(M1+m2+m3)

knowing a, then the tension between m2 and m3 is

Tension=m3*a