usy02altheexpertta.com watch Hand shakers Episode to Kitten Kachen scientino Not
ID: 1546809 • Letter: U
Question
usy02altheexpertta.com watch Hand shakers Episode to Kitten Kachen scientino Notason To oecina Notation cow Class Management Torque: Moment ofinertia; Rotational Dynamics Begin Date: 3/6/2017 12-00:00 AM Due Date: 320/2017 50000 PM End Date: 3 3/2017 12 00:00 AM (8%)Problem You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M.length L)that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achieve static equilibrium even if they are of different mass, as most people are. You have decided that each rider will be positioned so that hisher center of mass will be a distance xoose from the end of the board when seated as shown. You have selected a child of mass m on the right), and an adult of mass n (shown times the mass of the child (shown on the left to test out your prototype. A20% Part (a) Derive an expression for the torque applied by the adult rider on the left in terms of given quantities and variables available in the palette. Assume counterclockwise is positive. Grade Summary I give deduction per feedback. Hintsals deduction perhiet. Hiens remaining 20% Part (b) Derive an expression the torque applied by the child rider on the righty in terms of given quandeles and variables for available in the Assume ise is positive. board in serms of given quantities and available inthe palerme. 20% palette. for the applied by the variables Part (e) Derive an expression and the masses and lengths in the problem. of n 4 20% Part (e) Determine the magnitude of the force exened on the pivot point by the see-saw while in use in torms of given quannee md in the palete. 888,4Explanation / Answer
Since the adult is on the left side of the pivot point, the adult’s weight will cause the board to rotate counter clockwise. When the pivot point is moved to the left of the center of the board, the weight of the board will be on the right side of the pivot point. This means board’s weight will cause the board to rotate clockwise. Since the child is on the right side of the pivot point, the child’s weight will cause the board to rotate clockwise. For the board to be balanced the counter clockwise torque must be equal to the clockwise torque. Let’s determine the weights.
For the adult, weight = n*mg
For the board, weight = Mg
For the child, weight = mg
To determine the torque, we must determine the distance from the pivot to each person. To determine this distance we need to the distance each person is from the center of the board. Since the board is L meters long, each half of the board is L/2 meters long. Since each person is xoffset meter from the board, each person is (0.5L - xoffset) meters from the center of the board.
The distance from the adult to the pivot point is (0.5L - xoffset – d)
Torque = nmg* (0.5L - xoffset – d)
The distance from the center of the board to the pivot point is d.
Torque = Mg* d
The distance from the child to the pivot point is (0.5L - xoffset + d)
Torque = mg*(0.5L - xoffset + d)
Total clockwise torque = Mg*d + mg*(0.5L - xoffset + d)
Set the torques equal to each other and solve for d.
nmg* (0.5L - xoffset – d) = Mg*d + mg*(0.5L - xoffset + d)
0.5*nmgL - nmg*xoffset - nmgd = Mg*d + 0.5mgL - mg*xoffset + mgd
Calculate d from here.
The magnitude of the force exerted on the pivot point by the see-saw while in use is equal to the sum of the weights.
F = nmg + Mg + mg = (nm + m + M)g
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.