Question

Figure 5 shows a crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is us = 0.25. (1) Determine the minimum horizontal force P required to hold the crate from sliding down the plane. (8 marks) Determine the minimum force P required to push the crate up the plane. (8 marks) (11)

Answer 1

Page 1 Ui) free body daigram [block tries to slide down) mocoso giren Ms=0.25 me 50 kg P > MSN 30 9=9,81 ml see = 30° mg sino 230° Evertical on Inclined plane=0 N= mg com o + p sin 30 p sin 30] 50x9.81 x 60130 + N = (404.7856 + 0.5P) & Holizantal forces on inclined plane =0 PcOs 30 + Ms N = mg sin 30 sin 30 Pcos 30 + 0.2 25[404.7856 +0.5p)= 50X9.51 0.991 p = 139.0536 P=140.316 Newtons minimum hâlizantal tölce p required to hold the cuate from sliding down the plane. CS Scanned with CamScanner

Pogi free body daửgram mg woso 30 MSN mg sino p.sin 30 + mg COA 30 Na = 434.7856 +0.5 p and & Haizantal taces on Inclined plane =0 PCOs30 = ng sin 30 + MSN P cos30 = 50X9.81 sin 30 +0.25 (424.7856+0.5p) 0.741 P = 351.4464 P= 474.2866 Newtons minimum boce required to push the crate up the plane. CS Scanned with CamScanner