Question

PART ONE

Answer 1

Answer:

25.7 N

44.8 N

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, buoyancy force ρVg pushing up, and normal force N pushing up.

Sum of forces in the y direction:

∑F = ma

N + ρVg − mg = 0

N = mg − ρVg

N = (3.5 kg) (9.8 m/s²) − (1000 kg/m³) (3.5 kg / 4000 kg/m³) (9.8 m/s²)

N = 25.7 N

The water pushes up on the block with a buoyancy force of ρVg.  According to Newton's third law, the block pushes back down on the water with an equal force of ρVg.

The other forces are weight force Mg pulling down, and normal force N pushing up.

Sum of forces in the y-direction:

∑F = ma

N − ρVg − Mg = 0

N = Mg + ρVg

N = (1.4 kg + 2.3 kg) (9.8 m/s²) + (1000 kg/m³) (3.5 kg / 4000 kg/m³) (9.8 m/s²)

N = 44.8 N