What is your weight? (in Newtons)

  • Thread starter Thread starter dalitwil
  • Start date Start date
  • Tags Tags
    Newtons Weight
Click For Summary
SUMMARY

The discussion centers on calculating the weight of a rider and their bike in Newtons, given specific parameters such as tire pressure and contact area. The gauge pressure in the tires is 73.5 lb/in², and the total contact area for both tires is 14.26 cm². The correct calculation shows that the rider's weight is approximately 654 N after accounting for the bike's weight of about 68.67 N. The final formula used is derived from the equilibrium of forces, equating tire pressure times total contact area to the combined weight of the bike and rider.

PREREQUISITES
  • Understanding of pressure units (e.g., lb/in² to N/m² conversion)
  • Basic knowledge of force and weight calculations (W = m*g)
  • Familiarity with area calculations for circular contact surfaces
  • Concept of equilibrium in physics
NEXT STEPS
  • Learn about converting pressure units, specifically from lb/in² to N/m².
  • Study the principles of equilibrium in static systems.
  • Explore calculations involving circular areas, particularly for tire contact surfaces.
  • Investigate the relationship between pressure, force, and area in fluid mechanics.
USEFUL FOR

Students in physics, engineers dealing with mechanical systems, and anyone interested in understanding weight calculations in relation to pressure and area.

dalitwil
Messages
23
Reaction score
0
Suppose that when you ride on your 7.00 kg bike the weight of you and the bike is supported equally by the two tires. If the gauge pressure in the tires is 73.5 lb/in2 and the area of contact between each tire and the road is 7.13 cm2, what is your weight? (in Newtons)

MY WORK:
P(gauge)=P-P(atmosphere)
73.5lb/in2=506765N/m2

P=506765N/m2+101000N/m2
=607765N/m2

F=P/A
=607765/.0713
=8.54e6N

This isn't correct. I also used the fact that the tires are round, so A=pi(d/2)^2 and used that as the area. That wasn't correct either.

I also tried multiplying (both forms of area) by 2, since there is 2 tires. Nope, not correct again.

And finally (all forms of force) I divided my answer by 9.81 since W does equal m*g. Still not correct.

I have been modifying my solutions in a variety of ways, none of which are correct, summerizing (no matter how many times i reread the section) I am not understanding what is going on at all.

Please help me.
 
Physics news on Phys.org
Dont forget to subtract the weight of the bike from the final force you get. The weight of the bike is about 70N. The total sufrace area of contact is 14.26cm^2.

You are dividing by only he surface area of one tire.
 
dalitwil said:
Suppose that when you ride on your 7.00 kg bike the weight of you and the bike is supported equally by the two tires. If the gauge pressure in the tires is 73.5 lb/in2 and the area of contact between each tire and the road is 7.13 cm2, what is your weight? (in Newtons)

MY WORK:
P(gauge)=P-P(atmosphere)
73.5lb/in2=506765N/m2

P=506765N/m2+101000N/m2
=607765N/m2

F=P/A
=607765/.0713
=8.54e6N

This isn't correct. I also used the fact that the tires are round, so A=pi(d/2)^2 and used that as the area. That wasn't correct either.

I also tried multiplying (both forms of area) by 2, since there is 2 tires. Nope, not correct again.

And finally (all forms of force) I divided my answer by 9.81 since W does equal m*g. Still not correct.

I have been modifying my solutions in a variety of ways, none of which are correct, summerizing (no matter how many times i reread the section) I am not understanding what is going on at all.

Please help me.
From problem statement:
{Weight of Bike} = (7 kg)*(9.81 m/sec^2) = (68.67 N)
{Rider's Weight} = W
{Total Tire Contact Area} = 2*(7.13 cm^2) = 2*(7.13e(-4) m^2)
{Tire Pressure Holding Bike + Rider} = (73.5 lb/in^2) = (5.0677e(+5) N/m^2)

Because system is at equilibrium, we have:
{Force Applied By Tire Pressure} = {Weight Bike + Rider}
::: ⇒ {Tire Pressure Holding Bike + Rider}*{Total Tire Contact Area} = {Weight Bike + Rider}
::: ⇒ (5.0677e(+5) N/m^2)*{2*(7.13e(-4) m^2)} = W + (68.67 N)
::: ⇒ (722.65 N) = W + (68.67 N)
::: ⇒ W = (654 N) ::: (or mass of 66.6 kg = 147 lbs)


~~
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
13K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
26K
  • · Replies 9 ·
Replies
9
Views
7K