Question

four blocks, with masses m1=5.9, m2= 5.50, m3=7.8, m4=7.4kg, are pulled on a horizontal frictionless surface by 43.0 n force that makes a 23 degree angle (theta) with the horizontal. what is the magnitude of the tension between the m3 and m4 blocks?

Answer 1

To find the magnitude of the tension between the m3 and m4 blocks, calculate the acceleration of the system first using the horizontal component of the applied force. Then, use the equation T = m3 * a to find the tension.

Step-by-step explanation:

To find the magnitude of the tension between the m3 and m4 blocks, we first need to calculate the acceleration of the system.

The 43.0N force can be divided into its horizontal and vertical components: Fx = F * cos(theta) and Fy = F * sin(theta).

Since the surface is frictionless, the net external force in the horizontal direction is equal to the mass times acceleration, so Fx = (m3 + m4) * a.

Now, we can substitute Fx into the equation and solve for the acceleration, a. a = Fx / (m3 + m4) = (43.0N * cos(23°)) / (7.8kg + 7.4kg) = 1.09 m/s².

Finally, the tension between m3 and m4 can be found using the equation T = m3 * a = 7.8kg * 1.09 m/s² = 8.51 N.