Determine the normal force of a crate on the floor

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SUMMARY

The normal force exerted on a 50kg crate being pulled across a level floor with a 200N force at a 30-degree angle is calculated to be 390N. The analysis involves understanding that the normal force is not equal to the gravitational force due to the vertical component of the applied force. By applying the equilibrium condition for vertical forces, the equation Fn + 100N - 490N = 0 is established, leading to the conclusion that Fn = 390N.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free body diagrams
  • Familiarity with trigonometric functions, specifically sine
  • Basic concepts of friction, including coefficients of kinetic friction
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dranseth
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Homework Statement


A boy pulls a 50kg crate across a level floor with an applied force of 200N. If this applied force is at an angle of 30 degrees with the horizontal and the coefficient of kinetic friction is .3, determine;
a) the normal force exerted on the crate by the floor.

I understand that the normal force =/= force of gravity in this case due to the applied force being at an angle. When I draw out the free body diagram, I get the following: (see the attachment)

I now know that I need to subtract the applied force vector that is perpendicular to the crate, but in my freebody diagram, both vectors are pointing up...?
 

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The resultant vertical force is zero.
That means that the sum of all the forces up is equal to the sum of all the forces down.
The forces up are the normal reaction and the component of the applied force. The force acting down is the weight. Can you now find the normal reaction?
 
Fnet = Fn +Fa - Fg = 0
Fn + 200sin30˚ - (50kg*9.8m/s/s) = 0
Fn + 100N - 490N = 0
Fn - 390N = 0
Fn = 390N

.. and that's the answer. Thanks a lot!
 

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