EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 4, Problem 26P
To determine
The experimental method to determine the force on one of the spring of the known weight of several persons and then use the method to determine the force constant of the car.
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A 29.0-kg box is released on a 28 degree incline and accelerates down the incline at 0.26 m/s.
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Chapter 4 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
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- Problem A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that you can obtain the coefficient of static friction Ps by measuring the critical angle e where slipping occurs. my Solution Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: EFx=mg Evaluating, we get: = mg But the frictional force is expressed as the product of coefficient and the normal force so. Us = mg For an inclined plane. The normal force is n=mg Then, Hs=mg cos(e) so tan(e)arrow_forwardWith this question do I need to add the y component of the applied force to the mg force or substract in this question. Can you also tell me when I do which operation?arrow_forwardThis figure shows a person preparing to do a pushup. Calculate the back extensor force (Fm) required to keep the trunk in the static position as shown. Report Fm, in units of N, to 0 decimal points (e.g. 101 N). Report only positive values (i.e. the absolute value). You are allowed an error margin of +/- 10 N from the exact correct answer. The spine and back extensor muscle (Fm) are aligned horizontally with the floor. The ground reaction force (GRF) under the hand is aligned vertically. You only need to consider the weight of the trunk (disregard the head and arms). Here is information you may need to solve this problem: • The GRF is 195 N. • The GRF has a moment arm of 0.4 m relative to the vertebra axis of rotation. • The weight of the trunk is 303 N. • W trunk has a moment arm of 0.29 m relative to the vertebra axis of rotation. • Fm has a moment arm of 0.06 m relative to the vertebra axis of rotation.arrow_forward
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- Please Asaparrow_forwardConsider the object subjected to the loading shown in the figure. Take the value of MD to be equal to 6 kN.m. 10 kN B 2 kN E 2 kN 2 m MD 2 m 2 m x Determine an equivalent force system at point A, using a vector approach. The value of F R of the equivalent force system is Ĵ) KN. The value of M RA of the equivalent force system is + k) kN.m. The magnitude of M RA is KN-m. (Round the final answer to four decimal places.)arrow_forwardWhich of the following statements are true? Select all correct. Question 4 options: An ideal pulley is frictionless. In a static-pulley system, the net force on individual pulleys may be non-zero. If a rope passes around many ideal pulleys, the tension may be different at either end of the rope. Using pulleys, it is possible to lift a box of weight mg using a tension T < mg If there is more than one rope in a pulley system, then each rope may have a different tension. If a rope passes over an ideal pulley, the tension is larger on one side than the other.arrow_forward
- Problem A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that you can obtain the coefficient of static friction Us by measuring the critical angle e where slipping occurs. Solution Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: EFx=mg Evaluating, we get: = mg But the frictional force is expressed as the product of coefficient and the normal force so. = mg For an inclined plane. The nomal force is n=mg Then, Hs=mg cos(e) so: tan(e)arrow_forwardProblem: A block is resting on a ramp as shown in the figure below. You can change the inclination angle θ by raising one end of the ramp. The block has a mass mb=6.5 kg. At the interface between the ramp and the block, the coefficient of static friction is μs=0.4, and the coefficient of kinetic friction is μk=0.28. I managed to calculate that: slip angle is θ= 21.8 degrees acceleration is 1.09 m/s^2. I am trying to find: Let the angle θ be the angle you calculated , θslip (21.8 degrees) . To keep the block from slipping, you apply a horizontal (parallel to the surface of Earth) force directed to the right on the block. Your horizontal force F→Yb has a magnitude of 45 N. Calculate the magnitude and direction of the force due to static friction for this new situation, assuming that the block does not slip. Enter a magnitude for this part, but consider what the sign of your answer tells you about the direction of the force.arrow_forwardProblem: A block is resting on a ramp as shown in the figure below. You can change the inclination angle θ by raising one end of the ramp. The block has a mass mb=6.5 kg. At the interface between the ramp and the block, the coefficient of static friction is μs=0.4, and the coefficient of kinetic friction is μk=0.28. I managed to calculate that: slip angle is θ= 21.8 degrees acceleration is 1.09 m/s^2. I am trying to find: 1. how long does it take from the time it slips to travel a distance d=0.95 m down the ramp?arrow_forward
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