# (a) The free-body diagram for the bureau is shown to the right. F is ...

(a) The free-body diagram for the bureau is shown to the right. F is ...

CHAPTER 6 SOLUTION FOR PROBLEM 3(a) The free-body diagram for the bureau is shown to the right.F is the applied force, f is the force of friction, F N is the normalforce of the floor, and mg is the force of gravity. Take the x axisto be horizontal and the y axis to be vertical. Assume the bureaudoes not move and write the Newton’s second law equations. Thex component is F − f = 0 and the y component is F N − mg =0.The force of friction is then equal in magnitude to the appliedforce: f = F . The normal force is equal in magnitude to theforce of gravity: F N = mg. As F increases, f increases untilf = µ s F N . Then the bureau starts to move. The minimum forcethat must be applied to start the bureau moving is...f.•...F Nmg...FF = µ s F N = µ s mg =(0.45)(45 kg)(9.8m/s 2 )=2.0 × 10 2 N .(b) The equation for F is the same but the mass is now 45 kg − 17 kg = 28 kg. ThusF = µ s mg =(0.45)(28 kg)(9.8m/s 2 )=1.2 × 10 2 N .

CHAPTER 6 SOLUTION FOR PROBLEM 21The free-body diagrams for block B and for the knotjust above block A are shown to the right. T 1 is themagnitude of the tension force of the rope pullingon block B, T 2 is the magnitude of the tension forceof the other rope, f is the magnitude of the force offriction exerted by the horizontal surface on block B,F N is the magnitude of the normal force exerted bythe surface on block B, W A is the weight of blockA, and W B is the weight of block B. θ (= 30 ◦ ) is theangle between the second rope and the horizontal.f......T 2 .F NT 1•.•.W BT 1........W AθFor each object take the x axis to be horizontal and the y axis to be vertical. The x componentof Newton’s second law for block B is then T 1 − f = 0 and the y component is F N − W B =0.The x component of Newton’s second law for the knot is T 2 cos θ − T 1 = 0 and the y componentis T 2 sin θ − W A = 0. Eliminate the tension forces and find expressions for f and F N in terms ofW A and W B , then select W A so f = µ s F N . The second Newton’s law equation gives F N = W Bimmediately. The third gives T 2 = T 1 / cos θ. Substitute this expression into the fourth equation toobtain T 1 = W A / tan θ. Substitute W A / tan θ for T 1 in the first equation to obtain f = W A / tan θ.For the blocks to remain stationary f must be less than µ s F N or W A / tan θ

CHAPTER 6 SOLUTION FOR PROBLEM 47The free-body diagram for the plane is shown to the right. F is the magnitudeof the lift on the wings and m is the mass of the plane. Since thewings are tilted by 40 ◦ to the horizontal and the lift force is perpendicularto the wings, the angle θ is 50 ◦ . The center of the circular orbit is to theright of the plane, the dashed line along x being a portion of the radius.Take the x axis to be to the right and the y axis to be upward. Thenthe x component of Newton’s second law is F cos θ = mv 2 /R and the ycomponent is F sin θ − mg = 0, where R is the radius of the orbit. Thefirst equation gives F = mv 2 /R cos θ and when this is substituted into thesecond, (mv 2 /R) tan θ = mg results. Solve for R:y....θ•xmgFR = v2g tan θ .The speed of the plane is v = 480 km/h = 133 m/s, soR =(133 m/s)29.8m/s 2 tan 50 ◦ =2.2 × 10 3 m .

CHAPTER 6 HINT FOR PROBLEM 19In each case you must decide if the block moves or not. If it moves the frictional force is kineticin nature; if it does not move the frictional force is static in nature. Assume the block does notmove and find the frictional force that is need to hold it stationary. Draw a free-body diagram forthe block. The forces on it are the gravitational force, the normal force of the plane, the frictionalforce of the plane, and the applied force P . Take the x axis to be parallel to the plane and the yaxis to perpendicular to the plane, then write the Newton’s second law equations in componentform, with the acceleration equal to zero. Calculate the frictional and normal forces and comparethe magnitude of the frictional force with the product of the coefficient of static friction and themagnitude of the normal force. If it is less the block does not move and the frictional force youcomputed is the actual frictional force. If the frictional force is greater the block does move andthe magnitude of the frictional force is the product of the coefficient of kinetic friction and themagnitude of the normal force.ans: (a) (17 N) î; (b) (20 N) î; (c) (15 N) îndans

CHAPTER 6 HINT FOR PROBLEM 29If the smaller block does not slide the frictional force of the larger block on it must have magnitudemg and this must be less than the product of the coefficient of static friction between the blocksand the magnitude of the normal force of the blocks on each other. Write the horizontal componentof the Newton’s second law equation for the smaller block. The horizontal forces on it are theapplied force and the normal force of the larger block. Do the same for the larger block. Theonly horizontal force on it is the normal force of the smaller block. According to Newton’s thirdlaw the two normal forces have the same magnitude. These equations can be solved for themagnitude of the normal force and then the value of the maximum static frictional force can becomputed.ans: 4.9 × 10 2 N

CHAPTER 6 HINT FOR PROBLEM 37To round the curve without sliding the frictional force must be equal to mv 2 /r, where m is themass of the bicycle and rider, v is their speed, and r is the radius of the curve, and it must beless than µ s F N , where µ s is the coefficient of kinetic friction between the tires and the road andF N is the normal force of the road on the tires.ans: 21 m

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