Torque - Lady on a diving board

  • Thread starter Thread starter tascja
  • Start date Start date
  • Tags Tags
    Board Torque
AI Thread Summary
A woman weighing 500 N is positioned at the end of a diving board measuring 3.90 m, supported by a fulcrum 1.40 m from the left end. To find the forces exerted by the bolt (F1) and the fulcrum (F2), the moments around the fulcrum can be calculated, leading to the equation F2 * 1.4 m = 500 N * 3.9 m. This allows for the calculation of F2 as 1392.86 N. Subsequently, F1 can be determined using the relationship F1 = F2 - 500 N. The solution effectively demonstrates the balance of forces on the diving board.
tascja
Messages
85
Reaction score
0

Homework Statement


A woman whose weight is w = 500 N is poised at the right end of a diving board with a length of Lw = 3.90 . The board has negligible weight and is bolted down at the left end, while being supported L2 = 1.40 m away by a fulcrum, as figure shows. Find Forces F1 and F2 that the
bolt and the fulcrum, respectively, exert on the board.

Homework Equations


T= rFsin(theta)

The Attempt at a Solution


Since there are two unknown forces, can i solve for them separately, assuming the axis of rotation is different in each?
to solve for F1 i would have that the fulcrum in the axis of rotation, assuming F2 is 0 for acting on the axis of rotation. and for F2 i would have the fulcrum at the bolt...
 

Attachments

Physics news on Phys.org
tascja said:

Homework Statement


A woman whose weight is w = 500 N is poised at the right end of a diving board with a length of Lw = 3.90 . The board has negligible weight and is bolted down at the left end, while being supported L2 = 1.40 m away by a fulcrum, as figure shows. Find Forces F1 and F2 that the
bolt and the fulcrum, respectively, exert on the board.
You can calculate momentuma around bolt.
ΣM=0 F2*1,4 - w*3.9=0
F2=w*3.9/1,4
After you get F2, you can easily calculate F1:F1=F2 - w
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top