22.8k views
1 vote
Two identical uniform planks of mass 3.5kg was set up as above. If the object on the left end of A has a mass of 2kg with density 2g/cm^3. Calculate the maximum value of x before the plank tilts over.

1 Answer

4 votes

The maximum value of x is 0.39 m.

Given information:

Mass of each plank (m1) = 3.5 kg

Mass of the object (m2) = 2 kg

Density of the object (d) = 2 g/cm³

Length of each plank (L) = 1 m

The volume of the object:

V = m2 / d = 2 kg / 2 g/cm³ = 1000 cm³

Convert volume to length (assuming object is rectangular):

l = V / (L * L) = 1000 cm³ / (100 cm * 100 cm) = 0.1 m

The total mass on the left side:

Mleft = m1 + m2 = 3.5 kg + 2 kg = 5.5 kg

The torque due to the left side (counterclockwise):

τleft = Mleft * g * x

Calculate the torque due to the right side (clockwise):

τright = m1 * g * (L - x)

Set the torques equal at the point of balance (maximum value of x):

τleft = τright

Mleft * g * x = m1 * g * (L - x)

5.5 kg * g * x = 3.5 kg * g * (1 m - x)

5.5x = 3.5 - 3.5x

9x = 3.5

x = 3.5 / 9

x = 0.39 m

User Georgio
by
8.4k points