Where is the centre of gravity for a person lying horizontally on two scales?

  • Thread starter Thread starter blueice2
  • Start date Start date
  • Tags Tags
    Equilibrium
AI Thread Summary
The discussion focuses on calculating forces and the center of gravity in two physics problems. The first scenario involves a fishing rod with a fish, requiring the determination of the force exerted by a child's hand and the ground forces, with given answers of 65.7N for both horizontal forces and 24.5N for vertical force. The second scenario examines a person lying on a board supported by two scales, with readings indicating the center of gravity is 0.76m from the head. Participants suggest using vector diagrams, resolving vectors into components, and applying Newton's Second Law to solve equilibrium problems. The conversation emphasizes the importance of showing effort and understanding the principles behind the calculations.
blueice2
Messages
2
Reaction score
0
Hi, I need some help with these questions:

A small fish [m=2kg] hangs from one end of a make-shift fishing rod [m=0.5kg] of length 2.5m. The other end of the fishing rod is braced against the ground and held horizontally by a child's hand, 1m from the bottom of the rod. Find the force exerted by the child's hand and the x and y forces on the rod from the ground. (Ans: 65.7N, 65.7N, 24.5N)

and

A 160cm tall person lies horizontally on a massless board supported by two scales. 1 is positioned at the person's feet and the other at the person's head. If the scale at head reads 32.8kg and the scale at the feet reads 29.4kg, where is the person's centre of gravity? (Ans: 0.76m from head)

Thanks
 
Physics news on Phys.org
You've got to show a little effort before someone's going to help you, but here's some general advice for any equilibrium problems.
1) Draw a vector diagram
2) Resolve all of your vectors into components, and draw a separate vector diagram with these.
3) Apply Newton's Second Law for each coordinate direction:
\sum F_{x}=0 , \sum F_{y}=0
4) Solve the equations you get for the unknowns.
 
So far for the first question, I did this:
Total Torque = 2kg(9.8)(2.5m) + 0.5(9.8)(1.25m)
I don't know how to find the distance of the Force Hand.
 
Centre of gravity

Assuming
1. the weight of the person consists of two parts, one at the end of the feet. the other at the end of the head.
2. the scales are put at the two edges, the feet and the head.
3. use the formula of centre of gravity.
 
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