An iron anchor weighs 250 pounds in air and has a weight density of 480 lbs/ft3. If it is immersed in sea water that has a weight density of 64 lbs/ft3, how much force would be required to lift it while immersed?
I know the answer is 217 pounds but I need help on trying to get that! Thanks <3
To find the force required to lift the iron anchor while immersed in sea water, we need to consider the difference in weight between the anchor in air and the anchor immersed in sea water.
First, let's find the weight of the anchor when immersed in sea water. The weight of an object can be calculated by multiplying its weight density with its volume.
The weight of the anchor in air is given as 250 pounds. The weight density of the sea water is 64 lbs/ft^3.
Let's assume the volume of the anchor is V cubic feet. Therefore, the weight of the anchor immersed in water is equal to the weight density of sea water multiplied by its volume:
Weight of anchor immersed in water = Weight density of sea water x Volume of anchor
Weight of anchor immersed in water = 64 lbs/ft^3 x V ft^3
Now, the weight of the anchor in air is equal to the weight of the anchor immersed in water plus the buoyant force acting on the anchor. The buoyant force is the weight of the water displaced by the anchor:
Weight of anchor in air = Weight of anchor immersed in water + Buoyant force
Since the anchor is fully submerged and in equilibrium, the buoyant force is equal to the weight of the water displaced by the anchor. The volume of water displaced is equal to the volume of the anchor (V ft^3). Therefore:
Weight of anchor in air = Weight of anchor immersed in water + Weight density of water x Volume of anchor
250 lbs = 64 lbs/ft^3 x V ft^3 + (64 lbs/ft^3 x V ft^3)
250 lbs = (128 lbs/ft^3 x V ft^3)
Now, we can solve for V, the volume of the anchor:
V ft^3 = 250 lbs / (128 lbs/ft^3)
V ft^3 = 250/128 ft^3
V ft^3 ≈ 1.9531 ft^3
Finally, we can substitute this value of V back into the equation for the weight of the anchor immersed in water:
Weight of anchor immersed in water = 64 lbs/ft^3 x V ft^3
Weight of anchor immersed in water = 64 lbs/ft^3 x 1.9531 ft^3
Weight of anchor immersed in water ≈ 125 lbs
Therefore, the force required to lift the iron anchor while immersed in sea water is approximately:
Force required to lift anchor = Weight of anchor in air - Weight of anchor immersed in water
Force required to lift anchor = 250 lbs - 125 lbs
Force required to lift anchor = 125 lbs
So, the force required to lift the iron anchor while immersed in sea water is approximately 125 pounds.
To determine the force required to lift the iron anchor while immersed in sea water, we need to consider the principle of buoyancy.
First, let's calculate the volume of the iron anchor. We know its weight density, which is the weight per unit volume, so we can use the equation:
Density (ρ) = Mass (m) / Volume (V)
Rearranging the equation, we can solve for volume:
Volume (V) = Mass (m) / Density (ρ)
The mass of the iron anchor is given as 250 pounds. Given its weight density of 480 lbs/ft^3, we can calculate the volume in cubic feet:
Volume (V) = 250 lbs / 480 lbs/ft^3
Volume (V) ≈ 0.521 ft^3
Now, let's calculate the buoyant force on the anchor when submerged in sea water. The buoyant force (Fb) is equal to the weight of the fluid displaced by the object, which follows Archimedes' principle.
The weight of the fluid displaced is equal to the weight density of the fluid times the volume of the fluid displaced:
Weight of fluid displaced = Weight density of fluid × Volume of fluid displaced
In this case, the weight density of sea water is given as 64 lbs/ft^3, and the volume of the iron anchor is 0.521 ft^3:
Weight of fluid displaced = 64 lbs/ft^3 × 0.521 ft^3
Weight of fluid displaced ≈ 33.344 pounds
Therefore, the buoyant force on the iron anchor while immersed in sea water is approximately 33.344 pounds.
Finally, to calculate the force required to lift the anchor, we need to subtract the buoyant force from the weight of the anchor:
Force required = Weight of anchor - Buoyant force
Force required = 250 pounds - 33.344 pounds
Force required ≈ 216.656 pounds
Rounding to the nearest pound, the force required to lift the iron anchor while immersed in sea water is approximately 217 pounds.
An aluminium bar weighs 17 kg in air. How much force is required to lift the bar while it is immersed in gasoline? The density of aluminium is 170 kg m-3 and that of gasoline is 42 kg m-3.
My answer:
Volume of aluminium=(mass of aluminium)/(density of aluminium)
=17/170=0.1〖 m〗^3(volume of aluminium is equal to volume of gasoline displaced)
Buoyant force = ρgV
= 42×10 × 0.1
= 42 N
Therefore, force required to lift = Actual weight – Buoyant force
= 17 ×10-42
= 170-42
= 128 N