161_PartUniversity Physics Solution

# 161_PartUniversity - Applying Newtons Laws Dividing the two equations tanstring = 5-33 a g sin slope g cos slope For the crate the component of the

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Applying Newton&s Laws 5-33 Dividing the two equations: slope string slope sin tan cos ag θ θ g θ + = For the crate, the component of the weight along the slope is cs l o p e sin mg θ and the normal force is l o p e cos . θ Using Newton&s Second Law again: c slope k c slope c sin cos θ θ ma μ −+ = . slope k slope sin cos θ g θ + = . 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 == ° ° = ° = . EVALUATE: In the limit that k 0 , string 0 and the string is perpendicular to the top of the crate. As k increases, string increases. Figure 5.75 5.76. IDENTIFY: Apply m F =a ! ! to yourself and calculate a . Then use constant acceleration equations to describe the motion. SET UP: The free-body diagram is given in Figure 5.76. EXECUTE: (a) y y Fm a ∑= gives cos nm g α = . x x a gives k sin mg α fm a −= . Combining these two equations, we have 2 k (sin cos ) 3.094 m s αμ =− = . Find your stopping distance: 2 0 0, 3.094 m s , 20 m s xx x va v = . 22 000 2 ( ) gives 64.6 m, x vv a =+ 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 ax x v = = . 00 0 2 ( ) gives 16 m s. x x v = EVALUATE: 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. IDENTIFY: Apply m F ! ! 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.) SET UP: 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. EXECUTE: 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 . AA A mg T = Block B (The only vertical forces on B are gravity down, and tension up): . BBB 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): ( ) . RB A R d T T m a L +−=

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5-34 Chapter 5 To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations : ( ) k k (/) () , o r . B RA ABR A B R mm d L m d mg mg m g m m m a a g L mmm μ +− −+ + = + + = ++ (a) When k 0, .
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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161_PartUniversity - Applying Newtons Laws Dividing the two equations tanstring = 5-33 a g sin slope g cos slope For the crate the component of the

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