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
Unformatted text preview: PHY 317K Homework 5 Solutions 1. D Since the initial forces are both 120 N, the weight of the plank is 240 N. This force acts at the center of gravity. After the support is moved to Z, the sum of vertical forces must still equal zero: F Y + F Z 240N = 0. Also, the sum of torques about any point must be zero. Choosing point Z (although any point is okay, and X would also be a natural point) for the torque: τ net = 3 4 LF Y 1 4 L (240N) = 0, where L is the length of the plank. Then: F Y = 1 3 (240N) = 80N. And F Z = 240N F Y = 160N. 2. C The center of mass of the rod is at its center. Any force acting at this point produces zero torque about the center of the rod. Since the forces acting on rod 1 and rod 2 produce a net torque about the center, these two rods cannot be put into static equi librium by an addition force acting at the center. However, there is no net torque on rod 3, so an addition (downward) force acting its the center can place rod 3 into static equilibrium. 3. C Define T to be the tension in one cable and F the horizontal force. The magnitude of F is 72 N. The weight of the child is mg = 145N. A freebody diagram of the forces acting on the child is useful. Simply require the sums of horizontal and vertical forces to be zero. mg F T 2 2 T x mg = 0, so T x = mg/ 2 = 0 . 5(145) N = 72 . 5N. 2 T y F = 0, so T y = F/ 2 = 0 . 5(72) N = 36N . Then T = T 2 x + T 2 y = (72 . 5N) 2 + (36N) 2 = 81N. 1 4. D The upward component of the tension T in one string is T sin θ . There are two strings, so the total upward force is twice this: 2 T sin θ 5. C We are interested in finding the horizontal force the wall exerts on the ladder. Notice that the 5 m ladder makes contact with the wall 4 m from the floor, since the distance from the wall to the bottom of the ladder is 3 m. (That is because 3 2 + 4 2 = 5 2 .) Since the ladder has negligible weight, the only forces on the ladder that produce non zero torque about the point where the ladder touches the floor are the man’s weight ( W = 720N) and the horizontal force of the wall. So, consider the net torque about the point of contact of the ladder with the floor: τ net = W (3 m) / 2 F wall (4 m) = 0. Then F wall = 3 W/ 4 = 1 . 5(720N) / 4 = 270N. 6. C We want F BD . The most direct way to get it is to consider the torque about the other end of the bar: τ net = F BD (1 . 6m) (16 N)(0 . 8m) (96 N)(0 . 4m) = 0 . Then F BD = 32N. 7. C The vertical forces on the bar must sum to zero, so F AC + F BD 16N 96N = 0 . Using the result of the previous problem for F BD : F AC = 16N + 96 N 32N = 80N . 8. D If T is the tension in the rope attached to the ceiling, T sin 30 ◦ 240N = 0, so T = 480N. The magnitude of the horizontal component of this tension is T cos 30 ◦ = . 866(480N) = 416N . The magnitude of the tension in the horizontal rope must also be 416 N, since those two forces must add to zero....
View
Full
Document
This note was uploaded on 11/19/2009 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.
 Spring '07
 KOPP
 Physics, Force, Gravity, Work

Click to edit the document details