A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point)the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance d0.233 m from therock, which has a mass of 325 kg, and fits one end of the rod under the rock's center of weight.If the homeowner can apply a maximum force of 703 N at the other end of the rod, what is the minimum total length L of the rodrequired to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner'sforce are both essentially vertical. The acceleration due to gravity is g9.81 m/s?.L = A young work crew is trying to quickly finish some touch-up painting to complete a job. Rather than setting up a secure work area,they decide to try a shortcut. Worker A will hold the far end of a 8.00 m work plank which weighs 115 N and is secured with ahinge on the opposite end. Worker B will stand on the plank 4.99 m from worker A. Worker B also brings up his paint can,m. =: 3.58 kg, which he places 0.250 m from where he is standing on the plank (between the two workers). Worker A unfortunatelyneeds to supply 509 N of force to hold the end of the plank. How much does worker B weigh? The acceleration due to gravityis g = 9.81 m/s².%3|weight of worker WB

Answered: Image /qna-images/answer/44b7f56a-37e3-439b-9f86-15118a6098b1.jpg

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance d = 0.233 m from the rock, which has a mass of 325 kg, and fits one end of the rod under the rock's center of weight. If the homeowner can apply a maximum force of 703 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical. The acceleration due to gravity is g = 9.81 m/s². L = m

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance d = 0.233 m from the rock, which has a mass of 325 kg, and fits one end of the rod under the rock's center of weight. If the homeowner can apply a maximum force of 703 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical. The acceleration due to gravity is g = 9.81 m/s². L = m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A beam resting on two pivots has a length of L = 6.00 m and mass M = 77.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 61.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (a) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. (b) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.x = (c) Check the answer to part (e) by computing torques around the first pivot point.x = (d)Except for possible slight differences due to rounding, is the answer the same for F and E?
The angle between the beam and the floor is 15.0 degrees. The angle between the rope and the horizontal is 31.0 degrees. The beam is 4.10-m long and has a mass of 6.10 kg. The box sits a distance of d = 0.770 m from the upper end of the beam and has a mass of 9.60 kg.What is the torque due only to the weight of the box if the axis is at the hinge?
The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center ( Ibar=ml² where l is the length of the bar). When the skater's hands and arms are brought in and wrapped around their body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. The hands and arms have a combined mass 10 kg. When outstretched, they span 2.1 m. When wrapped, they form a cylinder of radius 21 cm. The moment of inertia about the rotation axis of the remainder of the body is constant and equal to 0.6 kg-m2. If the original angular speed is 0.5 rev/s, what is the final angular speed? 00f= rev/s Determine how much the figure skater's rotational KE increased. [NOTE: We express energy in J. To get J, you must convert your angular velocities to the appropriate mks. units which are rad/s.] ΔΚΕ- =
(a) A woman opens a 1.35 m wide door by pushing on it with a force of 44.5 N applied at the center of the door, at an angle perpendicular to the door's surface. What magnitude of torque (in N · m) is applied about an axis through the hinges? (b) A girl opens the same door, using the same force, again directed perpendicular to the surface, but now the force is applied at the edge of the door. What magnitude of torque (in N · m) is applied about the axis through the hinges now?
A beam resting on two pivots has a length of L = 6.00 m and mass M = 87.0 kg. The pivot under the left end exerts a normal force n₁ on the beam, and the second pivot placed a distance = 4.00 m from the left end exerts a normal force n₂. A woman of mass m = 52.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. -L- m M (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin. (Submit a file with a maximum size of 1 MB.) Choose File No file chosen (b) Where is the woman when the normal force n₁ is the greatest? x = L m (c) What is n, when the beam is about to tip? N (d) Use the force equation of equilibrium to find the value of n₂ when the beam is about to tip. N (e) Using the result of part (c) and the torque equilibrium equation, with torques computed…
The wheels, axle, and handles of a wheelbarrow weigh 60.0 N. The load chamber and its contents weigh 525 N. The drawing shows these two forces in two different wheelbarrow designs. To support the wheel- barrow in equilibrium, the man’s hands apply a force F to the handles that is directed vertically upward. Consider a rotational axis at the point where the tire contacts the ground, directed perpendicular to the plane of the paper. Find the magnitude of the man’s force for both designs.
I need help with problem over Rotational Kinectic (Equilibrium).. A window washer weighing 825.6 N is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 262.5 N and is 4.034 m long. What is the force in each rope when the window washer stands 1.48 m from one end? Force in Rope 1 (Rope closest to the window washer.) = Force in Rope 2 (Rope farthest from the window washer.) =
A 5.2-m ladder rests against a wall as shown. The ladder has a mass of 27.6 kg. A 56.3-kg person stands on the ladder at a distance of 3.6 m from the bottom of the ladder. The foot of the ladder is 1.9 m from the bottom of the wall. Assume there is no friction between the ladder and the wall so that the force of the wall on the ladder is acting only in the horizontal direction. However, there is a friction force between the ladder and the ground. 1.) What is the force, in N, exerted by the wall on the ladder?
Consider a teeter-totter that is made out of pine, shown below. A pineapple sits on one side of the teeter-totter, a distance 1.38 m from the pivot point, while an apple (non-pine) sits on the other side, a distance of 2.65 m from the pivot. If the pineapple has a mass 125 g, what must the mass of the apple be in order to balance the teeter-totter (i.e. to maintain rotational equilibrium)? [Answer in units of grams (g) with 3 sig figs but do not enter units with your answer]
To get up on the roof, a person (mass 87.0 kg) places a 5.60 m aluminum ladder (mass 14.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom? top N bottom N
Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.0° above the horizontal. One child, with a mass of 41.0 kg, is hanging 1.05 m along the branch from the tree trunk. The other child, with a mass of 31.0 kg, is hanging 2.35 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.
Three children are trying to balance on a seesaw. The seesaw is 4.1 m long and pivots about a rock at its center. Louie and Laura are already sitting on opposite ends. If Martha weighs 33 kg, where should she sit so as to balance the seesaw?