Find the acceleration of a crate if the applied force is 330N and the coefficient of kinetic friction is .45. The crate has a mass of 32kg and you are pushing on it at a 21 degree angle. I'm not sure if this part is important but the coefficient of static friction between the crate and floor is .57. Can you please show me how to do it.
Thanks :)
2006-10-23 13:14:23 · 1 answers · asked by bluevolleyball12 1 in Education & Reference ➔ Homework Help
When one surface is sliding over the other, the friction force between them is always the same, and is given by the product of the coefficient of kinetic friction and the normal force: F = μkN
I will assume that the 21 degree angle is pushing slightly upward, which means that we are applying a force of Sin(21)*330N upward, and a force of Cos(21)*330N horizontally.
The crate weighs 32kg and is also being acted on by gravity in the vertical direction downward.
This force is 32*9.8=313.6N
So the acceleration is what's left over when you subtract the the frictional force and divide by the mass
The frictional force is:
.57 * (313.6-Sin(21)*330)
=111.343N
So the acceleration is
(cos(21)*330-111.343)/32
=6.15m/s^2
j
2006-10-25 07:22:22 · answer #1 · answered by odu83 7 · 0⤊ 0⤋