phys124s11-hw02

phys124s11-hw02 - HW 2 Solutions Test Your Understanding...

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1 HW 2 Solutions Test Your Understanding 11.1: Conditions for Equilibrium A drum major marching in a parade tosses his baton in the air. As the baton rises and falls, it rotates at a uniform rate. Which condition(s) for equilibrium does this motion satisfy? the first condition for equilibrium neither the first condition nor the second condition for equilibrium the zero condition for equilibrium the second condition for equilibrium both the first condition and the second condition for equilibrium The baton has a downward linear acceleration due to gravity, so the first condition for equilibrium (that the net force on the object is zero) is not satisfied. The baton rotates at a constant rate and so has zero angular acceleration, so the second condition for equilibrium (that the net torque on the object is zero) is satisfied. Torques on a Ruler. A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. One coin, of mass M 1 = 10 g, is placed at the zero mark; the other, of unknown mass M 2 , is placed at the 4.7 cm mark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is perfectly balanced. Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot? Yes/No Note that the weight of the ruler itself also does not exert a torque with respect to the pivot point since the rule is uniform and is pivoted at its midpoint. Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. Why not? Because the distance from the exerted force to the pivot is zero. Find the mass M 2 . Second condition for equilibrium: 0 i i τ = 11 2 2 2 2 30 10 3 1.7 1.7 Mgl Mgl g cm M cm M g ⋅= = ⋅ = Exercise 11.6 Two people are carrying a uniform wooden board that is 3.0 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.
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2 I DENTIFY: Apply the first and second conditions of equilibrium to the board. SET UP: The free-body diagram for the board is given in Figure 11.6. Since the board is uniform its center of gravity is 1.50 m from each end. Apply 0 y F = , with y + upward. Apply 0 τ = with the axis at the end where the first person applies a force and with counterclockwise torques positive. EXECUTE: 0 y F = gives 12 0 FFw +−= and 21 160 N 60 N 100 N FwF =− = = . 0 = gives 2 (1.50 m) 0 Fx w −= and 2 160 N 2.40 m 100 N w x F ⎛⎞ == = ⎜⎟ ⎝⎠ . The other person lifts with a force of 100 N at a point 2.40 m from the end where the other person lifts. EVALUATE: By considering the axis at the center of gravity we can see that a larger force is applied by the person who pushes closer to the center of gravity.
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phys124s11-hw02 - HW 2 Solutions Test Your Understanding...

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