A uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest

30 kg

Explanation:

Given that a uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest ?

Taking the moment at the middle

Sum of Clockwise moment = sum of anticlockwise moments

Sum of clockwise moment will be:

30 × ( 4 - 1 ) = 30 × 3 = 90Nm

At equilibrium, the sum of anticlockwise moments will be equal to sum of clockwise moment.

90 = 30 × ( 5 - 4 ) + M

90 = 30 × 1 + M

90 = 30 + M

M = 90 - 30

M = 60

Moment = W × distance

Let the minimum distance = 2

60 = 4W

W = 60/2

W = 30 kg

Therefore, the maximum mass will be 30kg

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it