The discussion revolves around calculating the maximum force required to push a 50 kg wagon up a 15-degree incline, which has a 15 kg box on top. The problem involves understanding the forces acting on both the wagon and the box, including friction coefficients and gravitational forces.
There is ongoing exploration of the problem with various interpretations of the forces involved. Some participants are attempting to clarify their reasoning and calculations, while others are questioning assumptions about the direction of forces and the application of friction. Guidance has been offered regarding the need to sum forces in different directions.
Participants express confusion about the application of forces and the role of friction, indicating a need for clearer definitions and understanding of the physics concepts involved. There is mention of potential gaps in knowledge from previous studies.
Fusilli_Jerry89 said:well first i drew it and came up with:
F-127-71+42.6=50a for the wagon and concluded that an extra 4.6 N must be directed down the slope in order to move the box. I am clueless now.
EthanB said:Sum the forces on the block in one free-body diagram, then sum the forces on the cart in another.
There should be no force of friction acting up the ramp on the wagon. The wagon's motion is upwards, and friction resists the direction of motion.
Once we have four equations from summing our forces, find a way to relate the maximum static friction between the block and cart to the force being applied to the cart. Plug in the maximum friction and get the maximum force.
Fusilli_Jerry89 said:What I don't get is how can you supply extra Newtons downwards without actually touch the box? The force of friction stays the same, this has to do with inertia but I have no idea how to solve for this? Maybe I missed something in Physics last year, could anyone help?