the force has an upward compnonet, which reduces weight.
fup=mg*sin45
So friction force then is
mu*mg(1-sin45)
so the horizontal component has to be equal to friction
Fcos45=mu*mg(1-sin45)
solve for F
What minimum force is required to drag a crate (m = 40 kg) across a floor (co-efficient of kinetic friction = 0.6) if the force is applied at a 45 degree angle upward from the horizontal?
Help ASAP please!!
2 answers
What is the answer and how do you get that answer?
2 answers