55 PM Sun Dec 18Substitute all known quantities.Solve for x substituting the normal force found inpart (b).webassign.net à-(75.7 kg)(9.80 m/s2)(2.46 m + x) + (57.8 kg) (9.80m/s²)(0) - (12.0 kg) (9.80 m/s2)(2.46 m) + n(2.46m) = 0-(1.82 x 10³ N-m) - (741.86)x - (289.296 N-m) +(2.46 m)n=0Exercise 8.3XRemarks The answers for x in parts (a) and (c) agree except for a small round-off discrepancy. This illustrateshow choosing a different axis leads to the same solution.mHints: Getting Started | I'm StuckSuppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. Asecond child sits at the end on the opposite side, and the system is balanced.(a) Find the mass of the second child.mchild 2kg(b) Find the normal force acting at the pivot point.N (a) Where should the man sit to balance theseesaw?Apply the second condition of equilibrium to theplank by setting the sum of the torques equal tozero.The first two torques are zero. Let x represent theman's distance from the pivot. The woman is at adistance L/2 from the pivot.Solve this equation for x and evaluate it.(b) Find the normal force n exerted by the pivot onthe seesaw.Apply the first condition of equilibrium to the plank,solving the resulting equation for the unknownnormal force, n.(c) Repeat part (a), choosing a new axis through theleft end of the plank.Compute the torques using this axis, and set theirsum equal to zero. Now the pivot and gravity forceson the plank result in nonzero torques.Substitute all known quantities.Solve for x substituting the normal force found inpart (b)..pivot + gravity+man+woman-00+ 0 Mgx + mg(L/2) - 0m (L/2)Mx 1.878=(57.8 kg) (2.46 m)75.7 kgm-Mg-mg - mp9 + n = 0n = (M + m + mpig= (75.7 kg + 57.8 kg + 12.0 kg) (9.80 m/s²)n= 1425.9N+pivot gravity + man + woman = 0-Mg(L/2 + x) + mg(0) - mp19(L/2) + n(L/2) = 0-(75.7 kg) (9.80 m/s2) (2.46 m + x) + (57.8 kg) (9.80m/s²) (0) (12.0 kg) (9.80 m/s2) (2.46 m) + n(2.46m) = 0-(1.82 x 10³ N-m) - (741.86)x (289.296 N-m) +(2.46 m)n = 0-

Length of the see saw = L = 4.92m Mass of the first child = m1=27.2 kg Mass of the second child = m2…

Exercise 8.3 Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. Mchild 2 kg (b) Find the normal force acting at the pivot point. N

Exercise 8.3 Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. Mchild 2 kg (b) Find the normal force acting at the pivot point. N

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 73P: A stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = = 4.00...

See similar textbooks

Related questions

Q: The particle of mass m = 2.2 kg is attached to the light rigid rod of length L = 0.80 m, and the…

Q: A brass wire is 1.40 m long and has a cross- sectional area of 6.00 mm². A small steel ball with…

Q: Insert Draw Design Layout V Tell me 2 Share Verdana v10 vA A A evA. Paragraph Styles te Dictat Ge n…

Q: Problem 11.69 • Part A A worker wants to turn over a uniform 1030-N rectangular crate by pulling at…

Q: F2 F1 0.180 m A square metal plate 0.180 m on each side is pivoted about an axis through point O at…

Q: H.W The uniform thin pole has a weight of 50 N and a length of 26m. If it is placed against the…

A: GivenWeight of pole ,W=50 NLength of pole AB ,L=26 md=10 mμs=0.3 at p[oint A,B and CForce P=150…

Q: A diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing…

Q: An aluminum rod is suspended by two wires, as shown in the figure below. The mass of the rod is 25.0…

Q: Part C In which case is less force required? O case A O case B

A: Given:- a bicycle wheel of mass m and radius R up over a curb of height h. To do this, you apply a…

Q: The diagram shows a top view of a child on a merry-go-round that is rotating counterclockwise.…

Q: A bridge of length 50.0 m and mass 8.10 x 104 kg is supported on a smooth pier at each end as shown…

Q: A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and…

A: Given data: A square metal plate is 0.180 m The given force is F1=24.0 NF2=16.7 NF3=14.6 N The…

Q: A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 58.0-gram…

Q: A dental bracket exerts a horizontal force of 66.2 N on a tooth at point B in the figure. What is…

Q: g automobile has a wheel base (the distance between the axles) of 2.80 m. The automobile's center of…

Q: HW1 A uniform disc of 220 mm diameter has a mass of 3.5 kg. It is mounted centrally in bearings…

Q: I Review | Constants A thin, uniform rod is bent into a square of side length a. Part A If the total…

A: Moment of inertia (I), also called angular mass of the body. It is the inertia of a rotating body…

Q: A 40 kg. 5.5-Im-long beam is supported, but not attached to, the two posts in the figure (Figure 1).…

A: Given:- mass of the beam= 40 kg mass of the man = 20 kg length of the beam is =5.5 m Find:- How…

Q: I Review | Constants A machine part has the shape of a solid uniform sphere of mass 255 g and…

Q: Determine the net torque on the 2.7-m-long uniform beam shown in the figure All forces are shown.…

A: It is given that the center of mass (CM) is at point C. So, the center of the rod is at point C. The…

Q: Review | Constants A uniform, 255 N rod that is 1.80 m long carries a 225 N weight at its right end…

Q: A monkey climbs a uniform ladder resting at an angle against a smooth wall and smooth floor. The…

Q: Two woights are connoctod by a vory lIght, Moxblo cord that passes over an 62.0 N frictionless…

A: Given: Mass m1 = 75.0 N. Mass m2 = 125 N. Mass of friction less pulley (M) = 62.0 N. Radius of…

Q: If the flywheel has a density (mass per unit volume) of 8600 kg/m, what thickness must it have to…

A: The disc starts rotating from rest with a uniform angular acceleration. Using newtons rotational…

Q: A centrifuge rotor rotating at 9500 rpm is shut off and is eventually brought uniformly to rest by a…

A: Solution:-Given thatInitial speed (Ni)=9500 rpmFrictional torque (T)=1.19 N.mmass of rotor (m)=3.67…

Q: Three children are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a…

Q: A pipe system consists of three pipes with lengths L₁ = 1.80 m, L₂ = 0.600 m, and L3 = 2.50 m. A…

A: We have for torque, τ→=r→×F→

Q: A solid uniform 3.25-kg cylinder, 65.0 cm in diameter and 12.4 cm long, is connected to a 1.50-kg…

A: mass of cylinder (M)=3.25 kgDiameter of cylinder (D)=65 cm=0.65 mlength of cylinder (L)=12.4…

Q: A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of…

A: Newton's Second Law of motion According to Newton's Second Law of motion, the rate of change of…

Q: A vaulter is holding a horizontal 3.00-kg pole, 4.50 m long. His front arm lifts straight up on the…

Q: A diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing…

Concept explainers

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

55 PM Sun Dec 18
Substitute all known quantities.
Solve for x substituting the normal force found in
part (b).
webassign.net à
-(75.7 kg)(9.80 m/s2)(2.46 m + x) + (57.8 kg) (9.80
m/s²)(0) - (12.0 kg) (9.80 m/s2)(2.46 m) + n(2.46
m) = 0
-(1.82 x 10³ N-m) - (741.86)x - (289.296 N-m) +
(2.46 m)n=0
Exercise 8.3
X
Remarks The answers for x in parts (a) and (c) agree except for a small round-off discrepancy. This illustrates
how choosing a different axis leads to the same solution.
m
Hints: Getting Started | I'm Stuck
Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A
second child sits at the end on the opposite side, and the system is balanced.
(a) Find the mass of the second child.
mchild 2
kg
(b) Find the normal force acting at the pivot point.
N

(a) Where should the man sit to balance the
seesaw?
Apply the second condition of equilibrium to the
plank by setting the sum of the torques equal to
zero.
The first two torques are zero. Let x represent the
man's distance from the pivot. The woman is at a
distance L/2 from the pivot.
Solve this equation for x and evaluate it.
(b) Find the normal force n exerted by the pivot on
the seesaw.
Apply the first condition of equilibrium to the plank,
solving the resulting equation for the unknown
normal force, n.
(c) Repeat part (a), choosing a new axis through the
left end of the plank.
Compute the torques using this axis, and set their
sum equal to zero. Now the pivot and gravity forces
on the plank result in nonzero torques.
Substitute all known quantities.
Solve for x substituting the normal force found in
part (b)..
pivot + gravity+man+woman-0
0+ 0 Mgx + mg(L/2) - 0
m (L/2)
M
x 1.878
=
(57.8 kg) (2.46 m)
75.7 kg
m
-Mg-mg - mp9 + n = 0
n = (M + m + mpig
= (75.7 kg + 57.8 kg + 12.0 kg) (9.80 m/s²)
n= 1425.9
N
+
pivot gravity + man + woman = 0
-Mg(L/2 + x) + mg(0) - mp19(L/2) + n(L/2) = 0
-(75.7 kg) (9.80 m/s2) (2.46 m + x) + (57.8 kg) (9.80
m/s²) (0) (12.0 kg) (9.80 m/s2) (2.46 m) + n(2.46
m) = 0
-(1.82 x 10³ N-m) - (741.86)x (289.296 N-m) +
(2.46 m)n = 0
-

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Angular speed, acceleration and displacement

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

What is the torque abot the origin of the force if it is applied at the point whose position is:
Under what conditions can a rotating body be in equilibrium? Give an example.
Check your Understanding Repeat Example 12.3 using the left end of the meter stick to calculate the torques; that is by placing the pivot at the left end of the meter stick.
Can a set of forces have a net force that is zero and a net torque that is not zero?
Find the net torque on the wheel in Figure P10.23 about the axle through O, taking a = 10.0 cm and b = 25.0 cm. Figure P10.23
Check Your Understanding A hollow cylinder is on an incline at an angle of 60 . The coefficient of static friction on the surface is s=0.6 . (a) Does the cylinder roll without slipping? (b) Will a solid cylinder roll without slipping?
What force must be applied to end of a rod along the x-axis of length 2.0 m in order to produce a torque on the rod about the origin of 8.0k Nm ?
When opening do you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. What torque are you exerting relative to the hinges?
Check Your Understanding Solve Example 12.1 by choosing the pivot at the location of the rear axle.
When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. How much torque are you exerting in newton-meters (relative to the center of the bolt)?
In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tee 15 m away as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?
A stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = = 4.00 m. A painter of mass m = 70.0 kg stands on the ladder d = 3.00 m from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE connecting the two halves of the ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately.