But what happens when the object is on an inclined surface?

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When a 30kg block is placed on a surface inclined at 30 degrees, the pressure it exerts can be calculated by determining the normal force. The normal force (Fn) is found using the equation Fn/Fg = cos(30 degrees), resulting in a normal force of 254.6N. This value is then divided by the area of force, which is 0.008m², yielding a pressure of approximately 3.2 x 10^4 Pa (Pascals). Understanding these calculations is essential for analyzing forces on inclined surfaces. The discussion effectively clarifies the process of calculating pressure in this context.
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But what happens when the object is on an inclined surface?

In this particuliar situation a 30kg block with an area of force of 80cm squared is standing on a surface which 30 degrees inclined relative to the floor. How do you calculate the pressure it exerts?
 
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You need to find the component of the force that is normal to the surface where it acts, in order to find the pressure.
 
Ok so Fn/Fg = cos(30deg) = 0.866
0.866 * 30kg * 9.8 = 254.6N
254.6 / 0.008m^2 = 3.2 X 10^4

Like this?
 
Good work, that's it! :smile:, btw N/m^2 = Pa (Pascal).
 

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