PHY183-Lecture32-1 - New Exam Seating Chart LAST NAME Ajayi...

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1 March 22, 2012 Physics for Scientists&Engineers 1 1 LAST NAME SIT IN ROW Ajayi -- Boss O Botros -- Cloonan N Coats -- Fetting M Finkbiner -- Harris L He -- Kasenow K Keller -- Maki J Marti -- Mindroiu I Moore -- Park H Patel -- Remington G Richardson -- Slater F Smith -- Trombley E Truong -- Xhaferllari D Xin -- Zilli C New Exam Seating Chart March 22, 2012 Physics for Scientists&Engineers 1 2 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 32 Examples Involving Static Equilibrium Student Standing on a Ladder Typically a ladder stands on the floor and leans against the wall Suppose a ladder of length = 3.04 m with mass m l = 13.3 kg, rests against a smooth wall at an angle of θ = 24.8 ° A student with mass 62.0 kg is standing on a rung that is r = 1.43 m along the ladder, measured from where the ladder touches the ground PROBLEM 1 What friction force must act on the bottom of the ladder to keep it from slipping? March 22, 2012 Physics for Scientists&Engineers 1 3 Student Standing on a Ladder SOLUTION Start with a free-body diagram The forces are R , the normal force exerted by the wall on the ladder N , the normal force exerted by the floor on the ladder W m , the weight of the student l , the weight of the ladder f s , the force of static friction between the floor and the ladder The student and ladder are in static equilibrium so March 22, 2012 Physics for Scientists&Engineers 1 4 F x , i i = 0 F y , i i = τ i i = 0 Student Standing on a Ladder The force components in the horizontal direction are The force components in the vertical direction are Sum the torques assuming that the pivot point is where the ladder touches the ground Solve for March 22, 2012 Physics for Scientists&Engineers 1 5 F x , i i = f s R = f s = R F y , i i = N m m g m l g = N = gm m + m l () i i = m l g 2 sin θ + m m g r sin R cos = 0 R = 1 2 m l g sin + m m g r sin cos = 1 2 m l g () + m m g r tan Student Standing on a Ladder Putting in our numerical values We found that s = so our answer is PROBLEM 2 Suppose that the coefficient of static friction between the ladder and the floor is 0.31. Will the ladder slip? SOLUTION 2 The normal force F was So the maximum static friction force is Which compares with s = 162 N, so the ladder will not slip March 22, 2012 Physics for Scientists&Engineers 1 6 R = 1 2 130. N + 608 N 1.43 m 3.04 m tan24.8 °= 162 N f s = 162 N N = m + m l = 9.81 m/s 2 62.0 kg + 13.3 kg = 738 N f s,max = μ s N = 0.31 738 N = 229 N
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2 Student Standing on a Ladder In general, the ladder will not slip if PROBLEM 3 What happens if the student climbs higher on the ladder? SOLUTION 3 Looking at the equation above, we can see that as the student climbs, r will increase until the force overcomes the maximum force of static friction Moral of the story • Don’t climb too high March 22, 2012 Physics for Scientists&Engineers 1 7 R = 1 2 m l g () + m m g r tan θ μ s m l + m m g
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This note was uploaded on 04/02/2012 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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PHY183-Lecture32-1 - New Exam Seating Chart LAST NAME Ajayi...

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