Cap 6 Do Halley - P06 016

Transcript

16. We choose +x horizontally rightwards and +y upwards and observe that the 15 N force has components Fx = F cos θ and Fy = −F sin θ. (a) We apply Newton’s second law to the y axis: N − F sin θ − mg = 0 =⇒ N = (15) sin 40◦ + (3.5)(9.8) = 44 in SI units. With µk = 0.25, Eq. 6-2 leads to fk = 11 N. (b) We apply Newton’s second law to the x axis: F cos θ − fk = ma =⇒ a = (15) cos 40◦ − 11 = 0.14 3.5 in SI units (m/s2 ). Since the result is positive-valued, then the block is accelerating in the +x (rightward) direction.