Question

Problem 1: A wood block is slid back and forth against a flat surface covered with sandpaper. The nominal contact area between the block and the paper is a rectangle with sides 0.03 m and 0.06 m in length. The block is slid back and forth 100 times. The hardness of the block is 5 MPa. This motion causes a 1.3 mm layer of material to be removed from the block's sliding surface. Use the diagram below for additional information. (a) What is the wear coefficient of the block against the sand paper? (b) Estimate the wear depth if the block is instead loaded to 200 N and slid back and forth 10 times? -100 N 0.5 m (sliding distance)

Answer 1

-125nio 29873 . o specific wear Rate - SWR= AW - Am 8XLXD SXLXD awt Weight difference L Load appered D Suding distance O I wear Coefficient Assancing The block is a small - block, so the density, s = 1000 kg/m² T = loon. De loox2D ._ DolX2X0.5 = loom SWR = Am - EXOV Jooo X100X100 SX (100) Am = 5.4V=SxAxh. L ES X 08% 0.03 X0.06X1:3x10 SWR = $.X 0.03 X0-06X/:38109 S (100) ² SWR 0.234 X10 am/s VEKFD = K.L.O FEL = normal force.. sliding distance. sliding dista D

Assmate Date Ir near volume - K V 0 LXD .03 0.06 X 1. 100 X 100 ke 0123481097 0.234 X109 K= LD. D. E = 200N D=0.5X10X2=lom V=KKLXD=0.234810 x 200 x 10. V= 0.468 x 10 m² Axh = 0.468 X 106 n= 0.4688105 0.4688106 - A 0.03X0-06 h = 260 x 156 m. Sh = 0.26 mm /