Question

Problem 6.8- Enhanced - with Feedback Part A To move a large crate across a rough floor, you push on it with a force F at an angle of 21 ° below the horizontal, as shown in the figure (Figure 1). You may want to review (LII pages 156-163) Find t floor i Expre Sub Figure 1 ▼of! 32 k Type here to search

Answer 1

The following figure shows the free body diagram of the box. Y-axis X-axis N F 217 F cos(21°) F sin (21°) mg The normal force on the box is given as follows: N = Fsin (21°)+ mg Therefore, the friction force is given as follows: f = UN = u(F sin (21)+mg) The horizontal force on the object is given as follows: F = F cos (21)

For box to move, the friction force should be less than the horizontal force on the box. The minimum force F can be calculated as follows: f = F, u Fsin (21°) + mg = F cos(21°) Rearrange for F. u(F sin( 21 )+mg = F cos (21° Fcos (21°)- uF sin (21°)= umg Fcos (21°) – u sin(21°))= umg umg F= cos (21°) – u sin( 21°) Therefore, the force, required is given as follows: F = (0.6)(32 kg) (9.8 m/s2 IN cos(21°)-(0.6 )sin (210) (1 kg.m/s2 = 261.857 N Therefore, the minimum force required is 262 N.