(a) The situation (b) Free-body diagram for toboggan w EXECUTE: The normal force has ouly a y-component, but the weight has both x-and y-components: Wx = w sina and Wy = -wcosa. (Compare to Example 5.4, in which the x-component of weight was -wsina. The difference is that the positive x-axis was uphill in Example 5.4, while in Fig. 5.12b it is downhill.) The wig-gly line in Fig. 5.l2b reminds us that we have resolved the weight into its components. The acceleration is purely in the + x-direction, so ay = O. New-ton's second law in component form then tells us that .l:.Fx = wsina = ma:x .l:.Fy = n -wcosa = may = 0 Since w = mg, the x-component equation tells us that mgsina = max' or Note that we didn't need the y-component equation to lind the acceleration. That's the beauty of choosing the x-axis to lie along the acceleration direction! What the y-components tell us is the magnitude of the normal force that the hill exerts on the toboggan: n = wcosa = mgcosa EVALUATE: Notice that the mass does not appear in our answer for the acceleration. This means that any toboggan, regardless of its mass or number of passengers, slides down a frictiouless hill with an acceleration of gsina. In particular, if the plane is horizon-tal, a = 0 and ax = 0 (the toboggan does not accelerate); if the plane is vertical, a = 90° and ax = g (the toboggan is in free fall). Notice also that the normal force n is not equal to the tobog-gan's weight (compare Example 5.4 in Section 5.1). We don't need this result here, but it will be useful in a later example.

CAUTION Common freHJody diagram elTOlS Figure 5.13 shows both the correct way (Fig. 5.13a) and a common incorrect way (Fig. 5.13b) to draw the free..body diagram for the toboggan. 'The diagram in Fig. 5.13b is wrong for two reasons: the normal force 5.2 Using Newton's Second law: Dynamics of Partides 147 must be drawn pe!peIIdicuIar to the surface, and there's no such thing as the "mil force." U you remember that "normal" means "perpen-dicular" and that ma is not itself a force, you'll be well on your way to always drawing correct free..body diagrams. 5.13 Correct and incorrect diagrams for a toboggan on a frictionless hill. (a) Correct free·body diagram for the sled y R1GHT!~ Nonna! force is perpendicular to the surface. It'. OK to drsw the acceleration vector /'" adjacent to (but not L . touching) the body. ~ "'IIGHT! X (b) Incorrect free-body diagram for the sled y nt ,,~ ..•. 1be quantity rna is ~ not a force. WRONG ~ r '" )!( Nonnal force is not vertical because the surface (which Is alongthe}C-axis) )()( ... WRONG ~'X is inclined. Two bodies with the same acceleration You push a 1.00-kg food tray through the cafeteria line with a con-stant 9.O-N force. As the tray moves, it pushes on a 0.50-kg carton of milk (Fig. 5.14a). The tray and carton slide on a horizontal sur-face that is so greasy that friction can be neglected. Find the accel-eration of the tray and carton and the horizontal force that the tray exerts on the carton.