The problem involves three boxes with given masses resting on a smooth horizontal surface, subjected to a horizontal force. The objective is to determine the contact force between two of the boxes as they accelerate together under the applied force.
Some participants have provided calculations for acceleration and forces acting on the boxes. There is mention of using free body diagrams to analyze the forces, indicating a productive direction in the discussion. However, there is no explicit consensus on the final contact force value.
Participants are encouraged to show their work and reasoning, particularly in isolating forces acting on individual boxes. The discussion reflects a collaborative effort to understand the dynamics of the system without providing direct solutions.
It's F_net = ma. Please show an attempt to find the acceleration of the 3 blocks as they move together. Then go from there.Student3.41 said:Homework Statement
Three boxes rest side-by-side on a smooth horizontal floor. Their masses are 1.45 kg, 3.10 kg, and 5.20 kg, with the 3.10 kg one in the center. A horizontal force of 24.0 N pushes on the 1.45 kg mass which pushes against the other two masses. What is the contact force between the 3.10 kg and 5.20 kg boxes?
Homework Equations
F=ma
The Attempt at a Solution
PhanthomJay said:It's F_net = ma. Please show an attempt to find the acceleration of the 3 blocks as they move together. Then go from there.
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this is correctStudent3.41 said:Ok, well I calculated the Total Mass = 9.75kg, then used the formula F=ma -> 24.0N/9.75kg=2.47m/s^2. I got the acceleration,
that's the NET force on M1i then used the same formula to find the original force on M1 = 3.57N,
You are not drawing Free Body Diagrams (FBD's). Look at block 3, the 5.2 kg block. Isolate it from the rest of the blocks and determine the forces acting on it in the x direction. There is only one force acting on it, the contact force from the middle block accelerating it forward. Calculate that force using Newton 2.then M1 on M2 = 7.63N which I assumed added together F_2-3=7.63+ -3.57 = 4.06N