A wooden float, of density 700 kg/m3, is 0.580mlong, 0.300m wide, and 0.078m thi

Question

A wooden float, of density 700 kg/m3, is 0.580mlong, 0.300m wide, and 0.078m thick, and floats on the surface of a body of water.

Part A

A lead weight is now attached to the bottom side of the float, sufficient to cause the float's top to be just level with the water's surface. Determine the mass of the lead weight.

Express your answer using two significant figures

I tried 4.07 and 3.7kg

I tried from (previous work of chegg answers)

volume of wooden float = .58* .3 * .078 = .013572 m 3

weight of wooden float = 700 * .013572 = 9.5 kg

volume of water displaced by it when the top is just level = .013572 m 3

weight of water displaced = 1000 * .013572 = 13.57kg

therefore, the weight of lead = weight of water displaced - weight of wooden float = 4.07

Explanation / Answer

Volume of wood = l*b*h = 0.58*0.3*0.078 = 0.013572 m^3

mass = volume*density = 0.013572*700 = 9.5 Kg

weight of wood W1 = mg = 9.5*9.8 = 93.1 N

buoyancy force = V*density of water*g

B1 = 1000* 0.013572*9.8 = 133 N

weight of lead = W2 = m*g = 9.8 *m N

density of lead = 11360 kg/m^3

Volume of lead immersed = m/11360

buoyancy force on lead = V*density of water *g

B2 = (m/11360)*1000*9.8 = 0.8626*m N

equating the forces

B1 + B2 = W2 + W1

133 + 0.8626*m = 93.1 + 9.8*m

8.9374 m = 133-93.1 = 39.9

m = 4.46 Kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.