Ch. 6 Solutions

# Ch. 6 Solutions - CHAPTER 6 Problem 3. A 3700-kg barge is...

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CHAPTER 6 Problem 3. A 3700-kg barge is being pulled along a canal by two mules, as shown in Fig. 6-59. The tension in each tow rope is 1100 N, and the ropes make 25 ° angles with the forward direction. What force does the water exert on the barge (a) if it moves with constant velocity and (b) if it accelerates forward at 0.16m/s 2 ? Solution The horizontal forces on the barge are the two tensions and the resistance of the water, as shown on Figure 6-59. The net force is in the x direction, so 2 T cos25 ° ! F res = ma x , since T 1 = T 2 = T . (a) If a x = 0, F res = 2(1100 N)cos25 ° = 1.99 kN. (b) If a x = 0.16 m/s 2 , F res = 1.99 kN ! (3700 kg)(0.16 m/s 2 ) = 1.40 kN. FIGURE 6-59 Problem 3 Solution. Problem 7. A block is launched up a frictionless ramp that makes an angle of 35º to the horizontal. If the block’s initial speed is 2.2 m/s, how far up the ramp does it slide? Solution The acceleration up the ramp is ! g sin35 ° , so the block goes a distance in this direction calculated from the equation v 0 x 2 ! 2 g sin 35 ° ( x ! x 0 ) = 0 . Thus, x ! x 0 = ( 2 . 2 m / s ) 2 = 2 ( 9 . 8 m / s 2 )sin 35 ° = 43 . 1 cm .

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CHAPTER 6 123 Problem 9. A 15-kg monkey hangs from the middle of a massless rope as shown in Fig. 6-60. What is the tension in the rope? Compare with the monkey’s weight. Solution The sum of the forces at the center of the rope (shown on Fig. 6-60) is zero (if the monkey is at rest), T 1 + T 2 + W = 0 . The x component of this equation requires that the tension is the same on both sides: T 1 cos8 ° + T 2 cos172 ° = 0, or T 1 = T 2 . The y component gives 2 T sin8 ° = W , or T = W = 2 sin 8 ° = 3 . 59 W = 3 . 59 ( 15 kg )( 9 . 8 m / s 2 ) = 528 N . FIGURE 6-60 Problem 9 Solution. Problem 11. A 10-kg mass is suspended at rest by two strings attached to walls, as shown in Fig. 6-62. Find the tension forces in the two strings. Solution The force diagram is superimposed on Fig. 6-62. Since the mass is at rest, the sum of the forces is zero, T 1 + T 2 + W = 0 , which is true for the x and y components separately, T 1 cos 45 ° ! T 2 = and T 1 sin45 ° ! W = 0. Solving for the magnitudes of the tensions, and substituting 98 N for the weight, we find T 1 = 2 98 N = 139 N , and T 2 = T 1 = = 98 N . FIGURE 6-62 Problem 11 Solution.