Download presentation

Presentation is loading. Please wait.

Published byAdrian Lathrop Modified about 1 year ago

1
Examples and Hints for Chapter 5

2
Problem 5.17 A light rope is attached to a block with mass 4 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 10 N. a) Draw two free body diagrams, one for the 4 kg block and one for the block with mass m. b) What is the acceleration of either block? c) What is the mass m of the hanging block?

3
Free Body Diagrams 4 kg T mg T

4
Use the sum of the forces

5
Problem 5.24 A box of bananas weighing 40 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 and the coefficient of kinetic friction is a) If no horizontal force is applied to the box and the box is at rest, how large is the frictional force exerted on the box? b) What is the magnitude of the frictional force if a monkey applies a horizontal force of 6 N to the box and the box is initially at rest? c) What minimum horizontal force must the monkey apply to start the box in motion? d) What minimum horizontal force must the monkey apply to keep the box moving at a constant velocity? e) If the monkey applies a horizontal force of 18 N, what is the magnitude of the friction force and what is the box’s acceleration?

6
Free body diagrams 40 N n 6 N f n-40=0 6-f=ma

7
Part a) if there is no motion, is there friction? No 40 N n 6 N f n-40=0 6-f=ma

8
Part b) f= s *n for static situation 40 N n 6 N f n-40=0 6-f=ma Therefore, n=40 and s =0.4 so f=0.4*40=16 N So 16 N is the magnitude no matter what force applied!

9
Part c) f= s *n for static situation 40 N n 6 N f n-40=0 6-f=ma Therefore, n=40 and s =0.4 so f=0.4*40=16 N So 16 N is the maximum static friction so this is what must be applied

10
Part d) f= k *n for moving situation 40 N n 6 N f n-40=0 6-f=ma Therefore, n=40 and k =0.2 so f=0.2*40=8 N So 8 N is the minimum force which must be applied to keep the box moving

11
Part e) f= k *n for moving situation 40 N n 18 N f n-40=0 18-f=ma Therefore, n=40 and k =0.2 so f=0.2*40=8 N m=40/9.8=4.08 kg 18-8=4.08*a a=2.45 m/s 2

12
Problem 5.61 Block A in the figure weighs 60 N. The coefficient of friction between the block and the surface on which it rests is The weight, w, is 12 N and the system is in equilibrium. a) Find the frictional force exerted on block A b) Find the maximum weight for which the system will remain in equilibrium 60 N 12 N 45 0

13
Free Body Diagrams 60 N 12 N 45 0 ThTh f 60 N n 12 N TvTv T 45 0 T*sin(45 0 ) T*cos(45 0 ),Note: at 45 0, Tsin(45 0 )=Tcos(45 0 )

14
From examination, T*cos(45 0 )=12 N =T*sin(45 0 ) ThTh f 60 N n 12 N TvTv T 45 0 T*sin(45 0 ) T*cos(45 0 ),Note: at 45 0, Tsin(45 0 )=Tcos(45 0 ) So T h =T v =12 N and since system in equilibrium, f=T v =12 N

15
From examination, T*cos(45 0 )=12 N =T*sin(45 0 ) ThTh f 60 N n 12 N TvTv T 45 0 T*sin(45 0 ) T*cos(45 0 ),Note: at 45 0, Tsin(45 0 )=Tcos(45 0 ) n-60 N=0 so n=60 N F= s *n=0.25*60=15 N Since T V =T h, then maximum weight Is 15 N

16
Hints Problem 5.16– Draw it! Find the sum of the forces parallel to the surface and set them equal to mass*acceleration Problem 5.18—Sum the forces and calculate max acceleration. Use my favorite kinematic formula (hint: it starts v f 2 =..) to solve for distance. In part b) consider only the first glider Problem 5.25– Remember constant speed implies that the acceleration=0 Problem 5.50– Look at the nice example on page 187 Problem 5.60—Use the tension in the rope to connect between the two blocks. In part a, see hint on on part b) downward motion implies friction is less. Problem 5.62—Consider the blocks as a single unit of weight w Problem 5.76– Find the maximum acceleration a s *g and use my favorite formula (see problem 5.18). Back calculate to find v 0

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google