Solving a Warehouse Puzzle: The 20 Kg Crate

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In summary, to get the crate to start sliding across the floor, a minimum horizontal force of 113.7 N must be exerted on it. To keep the crate moving at constant velocity, a force of 310 N is needed. The coefficient of kinetic friction is needed to solve for the constant velocity, but is not given in the problem.
  • #1

Homework Statement


A 20 Kg Wooden crate rests on the wooden floor of a warehouse. What is the minimum horizontal force which must be exerted on the crate to get it to start sliding across the floor? Once it starts to move, how much force is needed to keep the crate moving at constant velocity?

m= 20 Kg
Ms=.58


Homework Equations


Ff=ma F=ma F=mg+ma
Ff=,58(20 x 9.8) 113.7=20a F=20(9.8)+20(5.7)
Ff=113.7 a=5.7 F=310N


The Attempt at a Solution



Above ^^. I do not know if my force is correct and how to find the constant velocity part!
 
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  • #2
ForceFysics said:

Homework Statement


A 20 Kg Wooden crate rests on the wooden floor of a warehouse. What is the minimum horizontal force which must be exerted on the crate to get it to start sliding across the floor? Once it starts to move, how much force is needed to keep the crate moving at constant velocity?

m= 20 Kg
Ms=.58


Homework Equations


Ff=ma F=ma F=mg+ma
Ff=,58(20 x 9.8) 113.7=20a F=20(9.8)+20(5.7)
Ff=113.7 a=5.7 F=310N


The Attempt at a Solution



Above ^^. I do not know if my force is correct and how to find the constant velocity part!
Welcome to PF, ForceFysics!

Your problem statement is missing the coefficient of kinetic friction.
If a body is at rest or moving at constant velocity, it has no acceleration, and thus, there is no net force acting on it (F_net = ma = 0). Use Newton 1 and the static friction coefficient when solving for part a, and use Newton 1 and the kinetic friction coefficient when solving for part b. The problem does not ask for the magnitude of the constant velocity, which is not solvable without more information.
 

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