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Homework Statement
If the 50-kg crate starts from rest and achieves a velocity of v = 4m/s m/s when it travels a distance of 5 mm to the right, determine the magnitude of force acting on the crate. The coefficient of kinetic friction between the crate and the ground is = 0.3
The crate is being pulled by the unknown force(p) at 30 degrees to the right.
Homework Equations
Fn=m*g
Ff=μ*Fn
∑F=m*a
a=dv/dt
v=ds/dt
The Attempt at a Solution
The weight ends up being 491.
I started off with summing up the forces in the x and setting them equal to m*a
∑Fx= cos(30)p-0.3(491)=50*a
Then I found the sum of the force in y and set it equal to 0.
∑Fy= sin(30)p+491=0
p=497
When I plug that back into the x equation I get a=5.7
Then I am not sure what to do next.
Do I use the a=dv/dt next?
Thanks again. I had statics last year and I am currently in dynamics and calc 1 so the integration, derivation, and sum of the forces no longer = 0 have me going crazy. I have never used this forum for help so I hope I set this up clearly for you.