$A$ force of $F = 10 \ N$ is exerted on a surface of area $A = 0.1 \ m^{2}$ at an angle of $60^{\circ}$ with the surface. Find the pressure produced on this surface.
(N/A) Pressure is defined as the normal force per unit area: $P = \frac{F_{\perp}}{A}$.
Given,the force $F = 10 \ N$ makes an angle of $60^{\circ}$ with the surface.
The component of the force perpendicular to the surface is $F_{\perp} = F \sin(60^{\circ})$.
$F_{\perp} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \ N$.
Given area $A = 0.1 \ m^{2}$.
Therefore,pressure $P = \frac{5\sqrt{3}}{0.1} = 50\sqrt{3} \ N/m^{2} \approx 86.6 \ N/m^{2}$.
Given,the force $F = 10 \ N$ makes an angle of $60^{\circ}$ with the surface.
The component of the force perpendicular to the surface is $F_{\perp} = F \sin(60^{\circ})$.
$F_{\perp} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \ N$.
Given area $A = 0.1 \ m^{2}$.
Therefore,pressure $P = \frac{5\sqrt{3}}{0.1} = 50\sqrt{3} \ N/m^{2} \approx 86.6 \ N/m^{2}$.
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