Magnitude of Normal Force (Frictionless)

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Homework Help Overview

The problem involves a crate being pushed across a frictionless horizontal surface with a force applied at an angle. Participants are exploring the calculation of the normal force acting on the crate, considering the effects of gravity and the applied force.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the crate, including gravity and the normal force. There is an exploration of the vertical components of the applied force and how they relate to the normal force. Questions arise about the third vertical component and its impact on the normal force calculation.

Discussion Status

The discussion is ongoing, with participants analyzing the forces involved and attempting to clarify the relationship between them. Some guidance has been provided regarding the components of the forces, but there is no explicit consensus on the final calculation of the normal force.

Contextual Notes

Participants are working within the constraints of a homework problem, focusing on the conceptual understanding of forces rather than providing a complete solution. The specific values for gravitational acceleration and the mass of the crate are noted, but the implications of these values are still being explored.

Spartan Erik
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Homework Statement



"A crate of mass 50 kg is pushed across a frictionless horizontal floor with a force of 100N directed 23.5 degrees below the horizontal. The magnitude of the normal force of the floor on the crate is:"

Homework Equations



F = ma? Not sure what else would apply.. mainly a conceptual issue here

The Attempt at a Solution



I took 100N x cos(23.5) in order to get the horizontal force which turns out to by 91.7N.. not sure what to do from here
 
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To solve for the normal force, analyze the vertical components of the forces acting on the crate. (Three forces act.)
 
Well gravity is acting on the crate (9.8 m/s^2 downward), and the crate itself has a normal force exerted in the opposite direction
 
Gravity and the normal force are two of the three forces. What's the third vertical component?
 
Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875
 
Spartan Erik said:
Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875
Right. So what must the normal force be to balance this force plus gravity?
 
Ah so 9.8 m/s^2 x 50kg = 490N

And adding that to 39.875N will result in a magnitude of 530N
 

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