![]() ![]() Again, try imagine it being a pressure tank. The top one is the force that the tank would exert on a scale, if it stood on one. ![]() If your tank is a pressure tank, that can be quite a lot. ![]() Example: Determine the amount of force exerted, in pounds, on 144 square. The bottom one calculates the total force applied on the bottom inside of the tank. This hydrostatic pressure can be determined with the formula: pressure force/area. For example, suppose that in a setting similar to the problem posed in Preview Activity 6. 2 ft 10 ft 2 from your text: A tank is 8 m long, 4 m wide, 2 m high and contains kerosene with density 820 kg/m3 to a depth of 1.5 m. Both You are calculating two different forces. What is the meaning of the value you find? Why?īecause work is calculated by the rule \( W = F \cdot d\), whenever the force \( F\) is constant, it follows that we can use a definite integral to compute the work accomplished by a varying force. Evaluate the definite integral \( \int^50_0 B(h) dh\). ft2 ft/ The ConstantDepth (object submerged horizontally Formula for Fluid Force: Fluid force Force per(unit area (.(b) Consider a vertical side of area 2 m × 1 m. Solution: We can find atmospheric pressure by either interpolating in. (a) Find the total force exerted by the water on the bottom surface on the tank. = B(h)\Delta h\) measure for a given value of \( h\) and a small positive value of \( \Delta h\)? Now sum forces normal and tangential to side AA. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |