a = f s,max /m = s g. 4. We first analyze the forces on the pig of mass m. The incline angle is.

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1 Chapter 6 1. The greatet deceleration (of magnitude a) i provided by the maximum friction force (Eq. 6-1, with = mg in thi cae). Uing ewton econd law, we find a = f,max /m = g. Eq. -16 then give the hortet ditance to top: x = v /a = 36 m. In thi calculation, it i important to firt convert v to 13 m/. 4. We firt analyze the force on the pig of ma m. The incline angle i. The +x direction i downhill. Application of ewton econd law to the x- and y-axe lead to mg in f ma mg co. Solving thee along with Eq. 6- (f = ) produce the following reult for the pig downhill acceleration: ag in co. To compute the time to lide from ret through a downhill ditance, we ue Eq. -15: 1 vt at t. a We denote the frictionle ( = ) cae with a prime and et up a ratio: t t / a / a a a 43

2 44 CHAPTER 6 which lead u to conclude that if t/t' = then a' = 4a. Putting in what we found out above about the acceleration, we have Uing = 35, we obtain =.53. gin 4g in co. 6. The free-body diagram for the player i hown to the right. i the normal force of the ground on the player, mg i the force of gravity, and f i the force of friction. The force of friction i related to the normal force by f =. We ue ewton econd law applied to the vertical axi to find the normal force. The vertical component of the acceleration i zero, o we obtain mg = ; thu, = mg. Conequently, f g 9.8 m/ THIK A force i being applied to accelerate a crate in the preence of friction. We apply ewton econd law to olve for the acceleration. EXPRESS The free-body diagram for the crate i hown to the right. We denote a the horizontal force of the peron exerted on the crate (in the +x direction), f i the force of inetic friction (in the x direction), i the vertical normal force exerted by the floor (in the +y direction), and mg i the force of gravity. The magnitude of the force of friction i given by Eq. 6-: f =. Applying ewton econd law to the x and y axe, we obtain repectively. f ma mg AALYZE (a) The econd equation above yield the normal force = mg, o that the friction i f mg g (9.8 m/ ) (b) The firt equation become mg ma which (with = ) we olve to find

3 45 a g m 55 g (.35)(9.8 m/ ).56 m/. LEAR or the crate to accelerate, the condition f mg mut be met. A can be een from the equation above, the greater the value of, the maller the acceleration under the ame applied force. 18. (a) We apply ewton econd law to the downhill direction: mg in f = ma, where, uing Eq. 6-11, f = f = = mg co Thu, with =.6, we have a = gin co = 3.7 m/ which mean, ince we have choen the poitive direction in the direction of motion (down the lope) then the acceleration vector point uphill ; it i decelerating. With v 18. m/ and x = d = 4. m, Eq. -16 lead to v v ad 1.1 m/. (b) In thi cae, we find a = +1.1 m/, and the peed (when impact occur) i 19.4 m/. 3. We ue the familiar horizontal and vertical axe for x and y direction, with rightward and upward poitive, repectively. The rope i aumed male o that the force exerted by the child i identical to the tenion uniformly through the rope. The x and y component of are co and in, repectively. The tatic friction force point leftward. (a) ewton Law applied to the y-axi, where there i preumed to be no acceleration, lead to in mg which implie that the maximum tatic friction i (mg in ). If f = f, max i aumed, then ewton econd law applied to the x axi (which alo ha a = even though it i verging on moving) yield co f ma co ( mg in ) which we olve, for = 4 and =.4, to obtain = 74.

4 46 CHAPTER 6 (b) Solving the above equation algebraically for, with W denoting the weight, we obtain W (.4)(18 ) 76. co in co (.4) in co (.4) in (c) We minimize the above expreion for by woring through the condition: d W (in co ) d (co in ) which lead to the reult = tan 1 = 3. (d) Plugging = 3 into the above reult for, with =.4 and W = 18, yield THIK In thi problem, the frictional force i not a contant, but intead proportional to the peed of the boat. Integration i required to olve for the peed. EXPRESS We denote the magnitude of the frictional force a v, where 7 m. We tae the direction of the boat motion to be poitive. ewton econd law give dv dv v m dt. dt v m Integrating the equation give v dv t dt v v m where v i the velocity at time zero and v i the velocity at time t. Solving the integral allow u to calculate the time it tae for the boat to low down to 45 m/h, or v v /, where v 9 m/h. AALYZE The integral are evaluated with the reult ln v t v m With v = v / and m = 1 g, we find the time to be m v m 1 1 g 1 t ln ln ln 9.9. v 7 /m

5 47 LEAR The peed of the boat i given by / v() t ve t m, howing exponential decay with time. The greater the value of, the more rapidly the boat low down. 41. Perhap urpriingly, the equation pertaining to thi ituation are exactly thoe in Sample Problem Car in flat circular turn, although the logic i a little different. In the Sample Problem, the car move along a (tationary) road, wherea in thi problem the cat i tationary relative to the merry-go-around platform. But the tatic friction play the ame role in both cae ince the bottom-mot point of the car tire i intantaneouly at ret with repect to the race trac, jut a tatic friction applie to the contact urface between cat and platform. Uing Eq. 6-3 with Eq. 4-35, we find = (R/T ) /gr = 4 R/gT. With T = 6. and R = 5.4 m, we obtain = The magnitude of the acceleration of the car a it round the curve i given by v /R, where v i the peed of the car and R i the radiu of the curve. Since the road i horizontal, only the frictional force of the road on the tire mae thi acceleration poible. The horizontal component of ewton econd law i f = mv /R. If i the normal force of the road on the car and m i the ma of the car, the vertical component of ewton econd law lead to = mg. Thu, uing Eq. 6-1, the maximum value of tatic friction i f,max = = mg. If the car doe not lip, f mg. Thi mean v g v Rg. R Conequently, the maximum peed with which the car can round the curve without lipping i v max Rg (.6)(3.5 m)(9.8 m/ ) 13 m/ 48 m/h.

Halliday/Resnick/Walker 7e Chapter 6

Halliday/Resnick/Walker 7e Chapter 6 HRW 7e Chapter 6 Page of Halliday/Renick/Walker 7e Chapter 6 3. We do not conider the poibility that the bureau might tip, and treat thi a a purely horizontal motion problem (with the peron puh F in the