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take a force, Fa 59N applied straight upward at the head A nail is stuck in a fl

ID: 1770524 • Letter: T

Question

take a force, Fa 59N applied straight upward at the head A nail is stuck in a floorboard so that it would of the nail, to pull it out. A carpenter uses a crowbar to pry the nail out, as shown in the diagram. The length of the handle is L-46 cm, and the length of the forked part, which grips the nail, is L 13 cm. Assume that the forked part is perfectly horizontal. The handle of the crowbar makes an angle -62° with the horizontal; the carpenter pulls in a horizontal direction. Eventually you will be asked to find the minimum force, Fpali that the carpenter must apply to pull the nail out 16. [2pt] Select all the statements that are correct while the carpenter pulls on the crowbar but before the nail begins to move. E.g., if A and C are correct, enter AC; if no statement is correct, enter N. You only have 4 tries A. Applying Fel parallel to the board as shown is the most effective way to remove the nail B. A possible lever arm in this problem has length L C. The sum of the forces on the nail is not zero. D. The line of action is parallel to Full and passes through the point where the carpenter's hand touches the handle. E. The carpenter must apyFit in order to remove the nail. Submit All AnswersLast Answer: abc Answer: Incorrect, ONE try left! 17. pt If a force Fpuil 12 N is applied to the crowbar as shown, what is the normal force of the board on the crowbar? Answer: Subit Al Answers 18. [lpt] What minimum force oull must the carpenter apply as shown in the diagram, to pull the nail out? Answer: Submit All Answers 19. [lpt] What is the minimum force Fpu required to pull the nail out if the carpenter can apply it at any angle relative to the handle? Answer: Submit All Answers

Explanation / Answer

16) bd

17)Balancing torque about nail,

N*Ln =Fpull*Lh* sin 62 degree

N =12*0.46/0.13* sin 62 degree

= 37.5 N

18) Fnail*Ln = Fpull*Lh* sin 62 degree

Fpull = 590*0.13/(0.46*sin 62 degree)

= 188.8 N

19)Fpull = 590*0.13/(0.46) = 166.7 N