PHY126 HW Set 3 (Chap.12)

Transcription

1 PHY16 HW Set 3 (Chap.1) Prblem 1.7 Descriptin: The figure shws a traffic signal, with the masses and psitins f its varius members indicated. The structure is munted with tw blts, lcated symmetrically abut the vertical member's centerline, as indicated. What tensin frce must the left-hand blt be capable f withstanding? 38 cm n l n r The frces n the traffic signal structure, and their lever arms abut pint O (n the vertical member s centerline between the blts) are shwn abve. The nrmal frce exerted by the blts and the grund n the vertical member are designated by n l and n r, measured psitive upward. The tw cnditins f static equilibrium needed t determine n l and n r are: 0 = ( Fy ) = nl nr (9.8 m/s )(30 kg 170 kg 65 kg) and = = )(0.38 m) 0 ( τ z ) ( nr nl -[(170 kg)(3.5 m) (65 kg)(8.0 m)](9.8 m/s ) These tw equatins can be rewritten as: 3 4 n n = N, n n = N r l r l 4 n l = Thus we find N, which is dwnward and must be exerted by the blt.

2 Prblem 1.30 Descriptin:A garden cart laded with firewd is being pushed hrizntally when it encunters a step 8.0 high, as shwn in the figure. The mass f the cart and its lad is 55, and the cart is balanced s that its center f mass is directly ver the axle. The wheel diameter is 60. F mg n R F s rcs =r(1-cs)=h rcs What is the minimum hrizntal frce that will get the cart up the step? Express yur answer using tw significant figures. We assume that a hrizntal push n the cart results in a hrizntal frce exerted n the wheels by the axle, as shwn. We als assume that bth wheels share the frces equally s that they can be treated tgether. Als shwn in the figure are the weight f the cart and the nrmal frce f the grund, bth acting thrugh the center f the wheels, and the frce f the step F s. If we cnsider the sum f the trques abut the step, the latter des nt cntribute, and the wheels (and cart) will remain statinary as lng as τ = MgRsin nrsin FR cs = 0 When n=0, hwever, the wheels begin t lse ( ). step cntact with the grund and g ver the step, s = cs F = ( N) tan(cs (1 h / R) = cs (11/15)) = 500 N is the minimum frce. (1 8/ 30). Then

3 Prblem 1.37 Descriptin: The figure shws a 66 sign hung centered frm a unifrm rd f mass 8. and length.3. At ne end the rd is attached t the wall by a pivt; at the ther end it's supprted by a cable that can withstand a maximum tensin f 800. What is the minimum height abve the pivt fr anchring the cable t the wall? Express yur answer using tw significant figures. Frm the equilibrium cnditin, τ = TLsin Mg( L / ) = 0, where T is the tensin n the cable, L is the length f the rd and M is ttal mass f the sign and the rd. Slving fr the tensin T = Mg /(sin ) = (1/ ) Mg 1 L / h, where we have used tan = h / L and the identity / 1 ct = 1/ sin. Slving fr h, we btain h = [(T /( Mg)) ]. Given the maximum tensin that culd be withstd in the cable, T max, the cnditin that must be met by h fr the sign t remain in static equilibrium is / Tmax (800 N) h = (.3 m) = 1.17 m. (66 kg 8, kg)(9.8 m/s ) S the Mg minimum height h min = 1.17 m.

4 Prblem 1.57 Descriptin:A unifrm ple f mass is at rest n an incline f angle, secured by a hrizntal rpe T f L n Mg as shwn in the figure. What is the minimum cefficient f frictin that will keep the ple frm slipping? Frm the equilibrium cnditins: 0 = F = f T cs Mg sin 0 = 0 = F // = n T sin mg cs τ = T ( L / )cs f ( L / ). The slutins fr the frce are f ( 1/ ) Mg sin, T = (1/ ) Mg tan =, and 1 sin n = Mg cs subject t the cnditin that f µn. Therefre cs sin tan sin µ cs µ. cs tan

5 Prblem 1.61 Descriptin: The figure shws a wheel n a slpe with inclinatin angle 0, where the cefficient f frictin is adequate t prevent the wheel frm slipping; hwever, it might still rll. The wheel is a unifrm disk f mass 1.50, and it is weighted at ne pint n the rim with an additinal mass. mg d R Mg Find the angle shwn in the figure such that the wheel will be in static equilibrium. Express yur answer using tw significant figures. The cw trque by the wheel abut the cntact pint is τ 1 = MgRsin. The ccw trque by the weight is τ = mgd = mgr(csφ sin ). Frm the equilibrium cnditin τ =. M m S ( M m)sin = mcsφ φ = cs sin. m 1 τ