Force Necessary to Start Moving Crate: 167 N

AI Thread Summary
To determine the force necessary to start moving a crate weighing 32 kg across a rough floor with a static friction coefficient of 0.57, the normal force must be calculated considering the angle of push at 21°. The weight of the crate is 313.92 N, but the angle affects the normal force, which requires resolving the applied force into its x and y components. The static friction force is calculated using the adjusted normal force, leading to confusion about how to incorporate the angle correctly. A diagram of the forces involved can help clarify the relationship between the applied force and the frictional force. Understanding these components is essential for accurately calculating the force needed to initiate movement.
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To move a large crate across a rough floor, you push down on it at an angle of 21°. Find the force necessary to start the crate moving, given that the mass of the crate is m = 32 kg and the coefficient of static friction between the crate and the floor is 0.57.

N=mg=(32)(9.81)= 313.92 * coefficaint of mew sub s=.57=178.9
Ok, I think I did that right...only I'm not sure what to do with the 21° angle. I did 178.9cos(21)=167.0 but that's not right.
 
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Did you draw a diagram with the applied force resolved into x and y components? (BTW, that's where the angle comes in)
Because in this case, N does not equal mg.
 
Like this?
Ny=N
Fky=0
Wy=-mgsin(21
Nx=0
Fkx=-mewk
Wx=0
I did Fk=mewkN=.57*(mgsin(21)=.57*112.5=64.12 I did somthing wrong...again...but I don't know what...
 
where's the force of the push? It has both an x and a y component.
But yeah, that's the sort of chart I had in mind. :grin:
 

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