exam2_review_torque

exam2_review_torque - Exercises on Statics and Rotational...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exercises on Statics and Rotational Dynamics Exercise 1.1 Jack, mass 50, Kg and Jill, mass 30 Kg, are sitting on a massless teeter-totter. Jack is 3 meters to the left of the fulcrum. How far away is Jill from the fulcrum? Since the teeter-totter is not rotating (or moving), the net torque about any point equals zero. Lets take the fulcrum as the point to evaluate the torque. Jills weight produces a clockwise torque about the fulcrum equal to 30 gx , where g is the acceleration due to gravity. Jacks weight produces a counter-clockwise torque about the fulcrum equal to 50(3) g about the fulcrum. Since the net torque must be zero, the clockwise and counter-clockwise torques must balance: 3(50) = 30 x x = 150 30 x = 5 meters Note that the calculation of the torques are easy in this case, since the force is per- pendicular to the radius vector. Exercise 1.2 A box, of mass 40 Kg, is placed at the end of a uniform plank. The plank has a mass of 80 Kg and a length of 10 meters. Where should the plank (plus box) be placed so that it balances on sharp fulcrum? That is, what is x in the figure? Since the plank (plus box) is stationary, the net torque about any axis must be zero. Lets choose the fulcrum as the point to evaluate the net torque. The box produces a counter-clockwise torque about the fulcrum of magnitude 40 x . We can treat the plank as if all the mass is located at the center, that is a mass of 80 Kg located at the center. Since the center of the plank is 5- x meters from the fulcrum, the plank produces a clockwise torque about the fulcrum of magnitude 80(5- x ). Since the net torque must be zero, the clockwise and counter-clockwise torques balance: 40 x = 80(5- x ) 40 x = 400- 80 x x = 3 . 333 meters 1 As in the last exercise, the torques were easy to calculate since the forces are perpen- dicular to the radius vectors. Exercise 1.3 Consider the following two vectors: ~ A : 5 units at 37 North of East ~ B : 10 units at 37 North of West Find ~ A ~ B . There are two ways to calculate the cross product: by using the unit vectors, or as | ~ A || ~ B | sin . In the latter case, the direction is determined using the right hand rule. In terms of the unit vectors, ~ A = 4 i + 3 j , and ~ B =- 8 i + 6 j . The cross product can be calculated using the determinant as shown in the figures page or ~ A ~ B = | ~ A || ~ B | sin = 5(10) sin (106 ) k = 48 k where k is a unit vector in the z-direction. Note that ~ B ~ A =- 48 k . Exercise 1.4 You want to build a sign that will hang at the end of a long rod. The sign has a weight of 50 pounds. The rod has a weight of 20 pounds and a length of 8 feet. One end of the rod is attached to the wall, and the other end is held up by a rope. The tension in the rope is labeled ~ T in the figure. Find the tension T in the rope and the force ~ F that the wall exerts on the rod....
View Full Document

This note was uploaded on 02/10/2010 for the course BIOL 202 taught by Professor Bloom during the Spring '10 term at University of North Carolina Wilmington.

Page1 / 30

exam2_review_torque - Exercises on Statics and Rotational...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online