Question

3 people are carrying a uniform metal plate in the shape of an isosceles triangle with base a = 0.7 m and height h = 1.35 m. Thickness of the plate is d = 1.5 cm, the density of the metal is ρ = 6500 kgm3. What is the upwards force that is exerted by the hands of each person, if each of them is holding the plate at a vertex and the plate is horizontal?

Answer 1

The upward force exerted by the hand of each person to support the metal plate is 150.54  N.

What is buoyant force?

The upward force exerted by the hands of the three people provides support on the metal plate.

The volume of air displaced by the metal isosceles plate is calculated as follows;

$V = A \times d\\\\V = \frac{1}{2} ba \times d\\\\V = \frac{1}{2} \times 0.7 \times 1.35 \times 0.015\\\\V = 0.00709 \ m^3$

The mass of the metal plate is calculated as follows;

$m = \rho V\\\\m = 6,500 \ kg/m^3 \ \times 0.00709 \ m^3\\\\m = 46.085 \ kg$

The upward force that must be provided by the 3 people to support the metal plate is calculated as follows;

$F_n = mg\\\\F_n = 46.085 \times 9.8\\\\F_n = 451.633 \ N$

The upward force exerted by the hand of each person to support the metal plate is calculated as follows;

$F_1 = F_2 = F_3 = \frac{F_n}{3} = \frac{451.633}{3} = 150.54 \ N$.

Learn more about upward forces here: https://brainly.com/question/6080367