Unless otherwise stated, all objects are located near the Earth\'s surface, wher
ID: 1594737 • Letter: U
Question
Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 { m m/s^2}. A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of 25 , or pull it at an upward angle 25 . If the crate has a mass of 53.0 kg and the coefficient of kinetic friction between it and the concrete is 0.790, calculate the required force to move it across the concrete at a steady speed for both situations.
Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 m/s2.
An object (mass 2.8 kg ) slides upward on a vertical wall at constant velocity when a force F of 57 N acts on it at an angle of 60 to the horizontal. Determine the force of kinetic friction on the object.
Explanation / Answer
The difference is that when he pushes down, he increases the normal force on the surface thereby increasing friction
in order to move the crate at constant speed, the horizontal pushing force must equal the force of friction; the horizontal moving force = F cos 25; the frictional force is u N where u is the coeff of friction and N is the normal force; we consider vertical equilibrium to find the normal force
if he pushes down, the vertical component of the pushing force is - F sin 25, and vertical equilibrium allows us to write
N - mg - F sin 25 = 0 or N = mg + F sin 25 = 519.4 + 0.42F
then, horizontal equilibrium gives us
F cos 25 - u N =0
F cos 25 - u(519.4 + 0.42F) =0 where u = 0.79
0.906F - 0.331F = 410.3N
F=713.5 N
if he pulls up, the normal force is given by
N - mg - F sin 25 =0
N=mg - 0.422F
and we have
F cos25 - u (mg + F sin 25) = F cos25 - u(519.4 + 0.42F ) =0
0.906F + 0.331F = 410.3N
F=331.4N
2) The force perpendicular to the wall is the horiz. component of the applied force (Fh)
Fh = 57.cos60 .. .. Fh = 28.5 N
= F / R
R = Fh = 28.5 N
F (due to friction) = [vert.comp. of applied force - mg] .. as object moves at constant velocity.
F = 57.sin60 - (2.8 x 9.80) .. .. F = 21.92 N
= F / R = 21.92 / 28.5 .. .. .. = 0.77
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.