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Question: A body of dimensions \((l \times b \times h)\) \(2m \times 3m \times 4m\) is acted upon by a force o...

A body of dimensions (l×b×h)(l \times b \times h) 2m×3m×4m2m \times 3m \times 4m is acted upon by a force of 50N50\,N on the top surface. Calculate the pressure due to force on that surface.

Explanation

Solution

Pressure is defined as the force acting per unit area. Mathematically it is expressed as
P=FAP = \dfrac{F}{A} where P is the pressure, F is the force applied and A is the area of cross section.
In this question, we need to first identify on which surface the force is acting and calculate its area. Since all the dimensions are different, the area in contact will be different for different sets of observations. After calculating the area in contact, we will simply put the values in the appropriate formula to get the answer.

Complete step by step solution:
It is being said that the force is acting on the top surface. So, we shall calculate its area.
The given dimensions are 2m×3m×4m2m \times 3m \times 4m where l=2m,b=3m,h=4ml = 2m\,,b = 3m\,,h = 4m .
Since it is the top surface mentioned in the question, the dimensions are l=2m,b=3ml = 2m\,,b = 3m\,
So, the area of the top surface is A=2×3A = 2 \times 3
A=6m2\Rightarrow A = 6\,{m^2}
Now the applied force is F=50NF = 50\,N
Substituting the known values in the formula of pressure,
P=FAP = \dfrac{F}{A} where P is the pressure, F is the force applied and A is the area of cross section.
P=506\Rightarrow P = \dfrac{{50}}{6}
P=8.33Nm2\therefore P = 8.33\,N\,{m^{ - 2}}

Note:
When the force is applied, it can be both a normal force or inclined to the object at some angle. When it is a normal force, we simply put in the value in the formula. However when the force is inclined we first calculate its perpendicular component better known as the thrust and then substitute in the formula.