Question

dynamics

Problem 3. Five forces pull on the 4.0 kg box in Fig. Find the net box's force (a) in unit-vector notation and (b) as a magnitude and direction. 14N SON 300 IIN 3.0N 4.0 kg hos 17 N

Answer 1

solution Five Forces pull on the 4.0kg box. BY 14 N s.ON 300 AN UN 3.0 4.0kg Box 17N a In unit-vector notation Newton's second low will give the box's acceleration, but we must First Find the sum of the forces on the box. Adding the a components of the forces gives. Z fx -IIN +14N COS 30° + 3.0N IIN +[14N X 0.866] +3.ON = - II N + 12:12 N +3.0N 4:124 N Two of the forces have only y componets Additing the y components of the fooces gives E fy + 5.0N + 14 N sin 30°-17N Page 1/3

= +50N + [14Nx0.5) -1711 5.ON 7N -17 N -5.0N so the Net Force on the box (in unit-vector notation) is ZF = (4.124N)2+(-5-ON Then we find the x andy components of the box's acceleration ceoing a = flm. E Fx Axe = (4-124 N) 1.031 m/s2 4.0kg elmis m Efn 1-5.0N) ау 5-1.2 mis? 4.0kg so in unit-vector form the acceleration of the box is (10 M/S2)i + (-1.2 m/s2); a = teor batedor by As a magnitude and direchon The acceleration found in port (a) has a magnitude of a = Tax²+ay² 2. JC1.0 m/s2)2+(-1.2 m/s2) Page 2/3

1-6 m/s² and to find its direction we calculele ау tan o age 1.2 1-0 Which gives us ton(-1-2) soº Here, and ax i's since ay is ve tve, lies in the proper this choice for a quadront. х Page 3/3