Lesson_4.4_Homework_Solution

Lesson_4.4_Homework_Solution - Lesson 18 Homework Solutions...

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Lesson 18 Homework Solutions 1. A 30-kg child sits on a seesaw 1.5 m from the pivot point. How far from the pivot point on the other side will his 45-kg friend has to sit for the seesaw to be in equilibrium? A is the pivot point. The weight of the 30-kg child is 294 N at a distance 1.5 m from the pivot. The weight of the 45-kg child is 441 N at a distance x from the pivot. 294 N A 1.5 m x 441 N The torque of 294 N about the pivot point A is: τ 1 = - 1.5 m × 294 N = - 441 mN (This is a clockwise torque) The torque of 441 N about the pivot point A is: τ 2 = x m × 441 N = 441x mN (This is a counter clockwise torque) When the seesaw is in equilibrium, the sum of the torques about A is zero. 441 x – 441 = 0 x = 1 m 2. A force of 36 N is applied to a particle located 0.22 m from its axis of rotation. What is the magnitude of the torque about this axis if the angle between the direction of the force and the torque arm is (i) 60 o , (ii) 90 o ? r = 0.22 m, F = 36 N (i) θ = 60 o The component of the force at right angles to the torque arm is 36 sin60 o . Torque about the pivot is: τ = r × F sin θ = 0.22 × 36 sin 60 o = 6.86 m N 0.22 m 60 o 36 N 36 sin 60 o (ii) θ = 90 o τ = r × F sin θ = 0.22 × 36 sin 90 o = 7.92 m N
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3. A uniform meter ruler is pivoted at its center with a 150 g mass suspended from the 30 cm mark. (i) At what position should a 100 g mass be placed to keep the system in equilibrium? (ii) What mass
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Lesson_4.4_Homework_Solution - Lesson 18 Homework Solutions...

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