1- A block of mass 3.0 kg lies on a horizontal surface. The cofficient of static

Question

1- A block of mass 3.0 kg lies on a horizontal surface. The cofficient of static friction between the block and the surface is 0.430. By raising one one end of the surface, the angle ( relative to horizontal ) of the surface is slowly incresed. At what angle ( in degrees) does that the block begin to slide down the inclined surface?

____________________________

2- A crate of mass 10.0 kg is being pushed by a person across a horizontal floor. The cofficient of static friction between the crate and the surface is 0.460, while the coefficient if kinetic friction os 0.230. The person starts to push, but the crate does not move. The person increases their pushing focre until suddenly it begins to move across the floor. what is the acceleration of the crate at the moment it begins to move ?

Explanation / Answer

Solution A

mgsin (theta) - (static) mu * mgcos(theta) = 0 rearrange the equation and cancal mg therefore, tan ( theta) = mu (static) theta = arctan (static mu) If the static coefficient is 0.43, then theta = arctan (0.57) theta = 23.26 degree Note: from the equation, the mass of the block is independent to the angle. Whether you have a bigger block or smaller block, it will start sliding @ 23.26 degree.

Solution 2

Fs = m*.565
Fd = m*.450

So the maximum static friction permissible between the crate and floor is 10*9.8*.46 = 45.08N
So that's also where Fd kicks in. where we ake kinetic friction
45.08N- m*.230 = ma
As you may notice,

ma =42.78

a=4.278 m/sec^2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.