Cap 6 Do Halley - P06 030

Transcript

30. Fig. 6-4 in the textbook shows a similar situation (using φ for the unknown angle) along with a free-body diagram. We use the same coordinate system as in that figure. (a) Thus, Newton’s second law leads to T cos φ − f = ma T sin φ + N − mg = 0 along x axis along y axis Setting a = 0 and f = fs,max = µs N , we solve for the mass of the box-and-sand (as a function of angle): 
T cos φ m= sin φ + g µs which we will solve with calculus techniques (to find the angle φm corresponding to the maximum mass that can be pulled). 
dm sin φm T =0 = cos φm − dt g µs This leads to tan φm = µs which (for µs = 0.35) yields φm = 19◦ . (b) Plugging our value for φm into the equation we found for the mass of the box-and-sand yields m = 340 kg. This corresponds to a weight of mg = 3.3 × 103 N.