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161_PartUniversity Physics Solution

# 161_PartUniversity Physics Solution - Applying Newtons Laws...

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Applying Newton°s Laws 5-33 Dividing the two equations: slope string slope sin tan cos a g θ θ g θ + = For the crate, the component of the weight along the slope is c slope sin m g θ and the normal force is c slope cos . m g θ Using Newton°s Second Law again: c slope k c slope c sin cos m g θ m g θ m a μ + = . slope k slope sin cos a g θ g θ μ + = . This leads to the interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic friction : k string tan tan(90 68 ) tan 22 0.40 θ μ = = ° − ° = ° = . E VALUATE : In the limit that k 0 μ , string 0 θ and the string is perpendicular to the top of the crate. As k μ increases, string θ increases. Figure 5.75 5.76. I DENTIFY : Apply m F = a ! ! to yourself and calculate a . Then use constant acceleration equations to describe the motion. S ET U P : The free-body diagram is given in Figure 5.76. E XECUTE : (a) y y F ma = gives cos n mg α = . x x F ma = gives k sin mg α f ma = . Combining these two equations, we have 2 k (sin cos ) 3.094 m s a g α μ α = = − . Find your stopping distance: 2 0 0, 3.094 m s , 20 m s x x x v a v = = − = . 2 2 0 0 0 2 ( ) gives 64.6 m, x x x v v a x x x x = + = which is greater than 40 m. You don°t stop before you reach the hole, so you fall into it. (b) 2 0 3.094 m s , 40 m, 0 x x a x x v = − = = . 2 2 0 0 0 2 ( ) gives 16 m s. x x x x v v a x x v = + = E VALUATE : Your stopping distance is proportional to the square of your initial speed, so your initial speed is proportional to the square root of your stopping distance. To stop in 40 m instead of 64.6 m your initial speed must be 40 m (20 m/s) 16 m/s 64.6 m = . Figure 5.76 5.77. I DENTIFY : Apply m F = a ! ! to each block and to the rope. The key idea in solving this problem is to recognize that if the system is accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn°t accelerate.) S ET U P : Take a positive coordinate direction for each object to be in the direction of the acceleration of that object. All three objects have the same magnitude of acceleration. E XECUTE : The Second Law equations for the three different parts of the system are : Block A (The only horizontal forces on A are tension to the right, and friction to the left): k . A A A m g T m a μ + = Block B (The only vertical forces on B are gravity down, and tension up): . B B B m g T m a = Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the portion of the rope that hangs vertically): ( ) .

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