×
Log in to StudySoup
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 9 - Problem 8p
Join StudySoup for FREE
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 9 - Problem 8p

Already have an account? Login here
×
Reset your password

A 140-kg horizontal beam is supported at each end. A

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 8P Chapter 9

Physics: Principles with Applications | 6th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

4 5 1 386 Reviews
26
5
Problem 8P

Problem 8P

A 140-kg horizontal beam is supported at each end. A 320-kg piano rests a quarter of the way from one end. What is the vertical force on each of the supports?

Step-by-Step Solution:
Step 1 of 3

SolutionStep1 of 4The impulse is large force acts on a body in short interval of timeWe know thatF p t pF =k t where kis proportional constant its value is 1 mvF = t because linear momentum p=mvF = mt here m is constantF t=mvF t=p where p=p -f iStep2 of 4J = p fp i J =mv- mu The graph is plotted force versus time, the area under the graph gives impulse\n Step 3 of 4 Given Mass of the tennis ball=m=60g=0.06 kg Initial velocity of the ball=u=32 m/s Final velocity of the ball=u=-32 m/s Change in momentum=p =To find the change in momentum we have the formulap=p -p f i =mv- mu =0.06×(-32)-0.06× 32 =-3.84 kg.m/s Now it is difficult find the area under the curve in above graph so split are into 2triangles and 1 rectangleArea of triangle 1= 2(Height)(base) = (F max)(t) 2 = 1(F )(2×10 )3 2 max 3 =F max ×10\nArea of triangle 2= 1(Height)(base) 2 = (F )(t) 2 max 1 3 = 2(F max)(2×10 ) 3 =F max×10Area of the rectangle 1= (Height)(base) = (F max)(t) = (F max )(2×10 )3 =2F ×10 3 max Now the sum of the area gives the impulseJ =Area of triangle 1+Area of triangle 2+Area of the rectangle 1 J = F max ×10 + F max ×10 3+ 2F max ×10 3Step4 of 4 According to impulse momentum theorem p=J-3.84 kg.m/s= F max×10 + F max ×10 3+ 2F max×10 3 3-3.84 kg.m/s= F max(4×10 ) 3.84 kg.m/sF max= 4×103F = -960 N maxSo the maximum value of the contact force during the collision is -960N And using the graph we can find maximum contact force easily

Step 2 of 3

Chapter 9, Problem 8P is Solved
Step 3 of 3

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

A 140-kg horizontal beam is supported at each end. A