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A block of iron of mass 7.5 kg and of dimensions 12 cm x 8 cm x 10 cm is kept on a table top on it's base of side 12 cm x 8 cm. Calculate (a) thrust and (b) pressure exerted on the table top. Take 1 kgf = 10 N.

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Answer

As we know,

(a) Force (F) = mass (m) x acceleration due to gravity (g)

Given,

m = 7.5 kg

1 kgf = 10 N

Substituting the values in the formula above we get,

F=7.5×10⇒F=75NF = 7.5 \times 10 \\[0.5em] \Rightarrow F = 75 N

Hence, F = 75 N

(b) As we know,

Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}

Given,

Area of the base = 12 x 8 = 96 cm2

Converting cm2 into m2

100 cm = 1 m

So, 100 cm x 100 cm = 1 m2

Hence, 96 cm2 = 1×9610000\dfrac{1 \times 96}{10000}

Therefore, A = 0.0096 m 2

Substituting the values in the formula above, we get,

P=750.0096⇒P=7812.5.5PaP = \dfrac{75}{0.0096} \\[0.5em] \Rightarrow P = 7812.5.5 Pa \\[0.5em]

Hence,

Pressure = 7812.5 Pa

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