- #1

physics565

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- Homework Statement
- John exerts a force of 10 N on a side of a table. The cross-sectional area is ##20 cm^2##. A friend of John exerts a force of 20 N on a different point of the same side of the table. The cross-sectional area is the same (##20 cm^2##). The two forces are of the same direction and perpendicular to the side of the table. Calculate the total pressure exerted on the table.

- Relevant Equations
- ##P=\frac{F}{A}##

I have come up with two different solutions but i'm not sure which is the correct one.

Pressure exerted by John:

##P_1=\frac{F_1}{A_1}=\frac{10}{0.002} pa=5000 pa##

Pressure exerted by John's friend:

##P_2=\frac{F_2}{A_2}=\frac{20}{0.002} pa= 10000 pa##

Total pressure:

##P=P_1+P_2=15000 pa##

Total Force: ##F=F_1+F_2=30 N##

Total area: ##A=A_1+A_2=0.004 m^2##

Total pressure: ##P=\frac{F}{A}=7500 pa##

**Solution 1**Pressure exerted by John:

##P_1=\frac{F_1}{A_1}=\frac{10}{0.002} pa=5000 pa##

Pressure exerted by John's friend:

##P_2=\frac{F_2}{A_2}=\frac{20}{0.002} pa= 10000 pa##

Total pressure:

##P=P_1+P_2=15000 pa##

**Solution 2**Total Force: ##F=F_1+F_2=30 N##

Total area: ##A=A_1+A_2=0.004 m^2##

Total pressure: ##P=\frac{F}{A}=7500 pa##