Question

5.2 Using Newton's Second Law: Dynamics of Particles 135 5.5 Our sketches for this problem. (a) Dirt-filled bucket pulls cart with granite block (d) Free-body diagram for cart Cart (c) Free-body diagram for bucket Bucket T W, sin 15 (b) Idealized model of the system Cart W 15 Bucket 15° W: cos 15 W2 EXECUTE: Applying F = 0 to the bucket in Fig. 5.5c, we find SOLUTION IDENTIFY and SET UP: The cart and bucket each move with a con- F, T(w) 0 Tw so stant velocity (in a straight line at constant speed). Hence each body is in equilibrium, and we can apply Newton's first law to each. Our target is an expression relating the weights wi and w. Figure 5.5b shows our idealized model for the system, and Figs. 5.5c and 5.5d show our free-body diagrams. The two forces on the bucket are its weight w and an upward tension exerted by the cable. As for the car on the ramp in Example 5.4, three forces act on the cart: its weight w, a normal force of magnitude n ex- erted by the rails, and a tension force from the cable. Since we're assuming that the cable has negligible weight, the tension forces ApplyingF = 0 to the cart (and block) in Fig. 5.5d, we get F T(w sin 150) 0 T= wi sin 15 so Equating the two expressions for T we find wisin 15=0.26w EVALUATE: Our analysis doesn't depend at all on the direction in which the cart and bucket move. Hence the system can moe with constant speed in either direction if the weight of the dirt and bucket is 26% of the weight of the granite block and cart. What would happen if w were greater than 0.26w? If it were less than 0.26w? Notice that we didn't need the equation F and block. Can you use this to show that n- wcos 15? that the cable exerts on the cart and on the bucket have the same magnitude T. (We're ignoring friction, so we assume that the rails exert no force on the cart parallel to the incline.) Note that we orient the axes differently for each body; the choices shown are the most convenient ones. We find the components of the weight force in the same way that we did in Example 5.4. (Compare Fig. 5.5d with Fig. 5.4b.) =0 for the cart

this example shows the uses of co-coordinate axes,

my question is why is it that we can substitute T for w2 for the second equation , won t it be wrong since we are using two different sets of axes

does it have something to do with expressing newtons second law in terms of magnittude?

can someone explain more about the uses of co-coordinate axes and when is it not apprioriate not to use it?

Answer 1

To this probem, the cart and the bucket are moving with constant velo city. dv O> a o V Constant- From NestDns law d motion EF =ma - EF- 0 Net force is żeyo 1ndicating eauiibrium of forces We have to find the yelation behween w, and w2 a cab le wshich bas a tension force T. Connected by free-bedy diagrams the cast and Draw the bucket. Choose appropriate co-odinate axes for eash the diagrams that can make the calculations easiex 니 Bucket T-WO T= W Now forget obout the axes taken for bucket take only eauation moving foroard cart T T-W,Sin Iso T 0.2C We -axis is taken parallef t6 the cart Here So, the nema force (N) is in ydirection which is eliminated in our caleulation From and T-W,D 26

Please review. Leave a comment if any doubt.