Dragging an object: what is Fmin and magnitude of acceleration?

  • Thread starter Thread starter tomiharv
  • Start date Start date
  • Tags Tags
    homework help
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 1K views
tomiharv
Messages
2
Reaction score
1
Homework Statement
By pulling a on rope attached to a crate of mass 124.0 kg, the crate is dragged across a floor. Let the coefficient of static friction between the crate and the floor be 0.860 and the angle between the rope and horizontal is 39.6 °. Find the minimum force required to move the crate.
At a later time the crate is now dragged with this same force, and has already started moving, on a wet floor where the kinetic friction coefficient is 0.516. What is the magnitude of the acceleration of the crate?
Relevant Equations
Fn = m*a or m*g (I think this is necessary)
I am not sure how to set this problem up mainly, I am unsure of what equations I need to be using right now. I have tried, for some dumb reason, a multitude of combinations with Cos(39.6) and 124 kg*9.8 m/s^2 as well as one time multiplying the coefficient. I am mainly just struggling with what equations I am meant to be using and believe me I have tried google and I just feel really stupid when I look at this problem so I don't know who else to ask as I feel bad milking my classmates for help with these problem set-ups.

I know that:

124*9.8 (m*g) = 1215.2 N
Cos (39.6) = 0.771 (rounded)
Sin (39.6) = 0.637 (rounded)

So I assume that Normal Force is the 1215.2 N right?

I also know that there's the applied force formula with x-component and y-component (Fa = Fax i + Fay j) would that be necessary for this problem?

I am going to set up a tutor for myself at my school to try and get more into the swing of knowing when to use what equations, there are just so many that I am a bit lost.
 
Physics news on Phys.org
I suggest starting with a nice free body diagram
 
  • Like
Likes   Reactions: tomiharv and PeroK
tomiharv said:
So I assume that Normal Force is the 1215.2 N right?
Wrong. ##mg=1215.2~##N is the weight of the crate. The pulling force has a vertical component away from the floor that lessens the burden of the full weight on the floor.

Follow @Gordianus's hint and draw a nice free body diagram. Then you will be able to see how this is put together.