{"id":1617,"date":"2018-03-23T13:11:26","date_gmt":"2018-03-23T05:11:26","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1617"},"modified":"2023-08-10T00:56:36","modified_gmt":"2023-08-10T00:56:36","slug":"buoyant-force-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/buoyant-force-problems-and-solutions.htm","title":{"rendered":"Buoyant force \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Buoyant force \u2013 problems and solutions<\/span><\/p>\n

1. A block of wood with length = 2.5 m, width = 0.5 m and height = 0.4 m. The density<\/a> of water is 1000 kg\/m3<\/sup>. If the block is placed in the water, what is the buoyant force<\/a>… Acceleration due to gravity is 10 N\/kg.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Volume of the block (V) = length x width x height = 2.5 x 0.5 x 0.4 = 0.5 m3<\/sup><\/span><\/p>\n

Density of water (\u03c1) = 1000 kg\/m3<\/sup><\/span><\/p>\n

Acceleration due to gravity <\/a>(g) = 10 N\/kg <\/span><\/p>\n

Wanted :<\/u> The magnitude of the buoyant force<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Formula of buoyant force :<\/span><\/p>\n

F = \u03c1 g V<\/span><\/p>\n

F = <\/i>buoyant force<\/i>, \u03c1 = <\/i>density of water<\/i>, g = <\/i>acceleration due to gravity<\/i>, V = volume<\/i><\/span><\/p>\n

F = (1000)(10)(0.5) = (1000)(5) = 5000 Newton<\/span><\/p>\n

2. Weight of an object in air is 100 N. The object is placed in a liquid. Increase in volume of liquid is 1.5 m3<\/sup>. If specific weight of the liquid is 10 N\/m3<\/sup>, what is the weight of the object in liquid.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Object’s weight<\/a> in air (w) = 100 Newton<\/span><\/p>\n

Increase in volume of liquid = volume of the object in liquid (V) = 1.5 m3<\/sup><\/span><\/p>\n

Specific weight of the liquid = 10 N\/m3<\/sup><\/span><\/p>\n

Wanted :<\/u> Object’s weight in liquid<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Object’s weight in liquid = object’s weight in air \u2013 buoyant force<\/span><\/p>\n

Object’s weight in liquid = 100 Newton \u2013 buoyant force<\/span><\/p>\n

Formula of buoyant force :<\/b><\/span><\/p>\n

FA <\/sub>= \u03c1 g V<\/span><\/p>\n

FA <\/sub>= buoyant force = the force exerted by the liquids on the object in water<\/span><\/p>\n

\u03c1 = density of liquid<\/span><\/p>\n

g = acceleration due to gravity<\/span><\/p>\n

V = object’s volume in liquid<\/span><\/p>\n

Specific weight :<\/b><\/span><\/p>\n

Specific weight of liquid = 10 N\/m3<\/sup><\/span><\/p>\n

w \/ V = 10 N\/m3<\/sup><\/span><\/p>\n

m g \/ V = 10 N\/m3<\/sup><\/span><\/p>\n

m (10) \/ V = 10 N\/m3<\/sup><\/span><\/p>\n

m \/ V = 1 kg\/m3<\/sup><\/span><\/p>\n

\u03c1 = 1 kg\/m3<\/sup><\/span><\/p>\n

The density of liquid is 1 kg\/m3<\/sup><\/span><\/p>\n

The magnitude of buoyant force <\/b>:<\/span><\/p>\n

FA <\/sub>= \u03c1 g V = (1 kg\/m3<\/sup>)(10 m\/s2<\/sup>)(1.5 m3<\/sup>) = 15 kg m\/s2 <\/sup>= 15 Newton<\/span><\/p>\n

Object’s weight in fluid :<\/b><\/span><\/p>\n

Object’s weight in fluid = 100 Newton \u2013 15 Newton<\/span><\/p>\n

Object’s weight in fluid = 85 Newton<\/span><\/p>\n

3<\/span>. <\/span>A ship sailing in the sea enters a wide and deep river. <\/span>T<\/span>he density of seawater is 1100 kg\/m<\/span>3<\/span><\/sup>, the <\/span>density of river water <\/span>is 1000 kg\/<\/span>m<\/span>3<\/span><\/sup>. <\/span>Determine comparison of the <\/span>volume of the object is in seawater and <\/span>in <\/span>river water.<\/span><\/span><\/p>\n

A. 11 : 10<\/span><\/p>\n

B. 10 : 11<\/span><\/p>\n

C. 121 : 100<\/span><\/p>\n

D. 1 : 1<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Density of seawater (\u03c11<\/sub>) = 1100 kg\/m3<\/sup><\/span><\/p>\n

Density of river water (\u03c12<\/sub>) = 1000 kg\/m3<\/sup><\/span><\/p>\n

Wanted:<\/u> Comparison of the <\/span>volume of the object is in seawater and <\/span>in <\/span>river water.<\/span> comparison of the <\/span>volume of the object is in seawater and <\/span>in <\/span>river water.<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

If <\/span>the object is <\/span>floating <\/span>t<\/span>hen buoyant force (F<\/span>B<\/span><\/sub>) = <\/span>weight <\/span>(w):<\/span><\/span><\/p>\n

\"Buoyant<\/span><\/p>\n

Archimedes<\/span>‘ principle<\/span><\/a> states that the buoyant force acting on an object in fluid<\/a> (water) is equal to the weight of the fluid (water) it displaces. The volume of the object in fluid (water) is equal to the volume of fluid (water) moved.<\/span><\/span><\/p>\n


\nComparison of the volume of the object in seawater and in river water:<\/span><\/span><\/p>\n

\"Buoyant<\/span><\/p>\n

The correct answer is B.<\/span><\/p>\n

4. Gold, whose mass is 193 grams is in kerosene having an upward force of 8000 dynes. If the <\/span>acceleration due to gravity <\/span>is 10 m\/s<\/span>2<\/span><\/sup> and the <\/span>density of <\/span>kerosene is 0.8 gr\/cm<\/span>3<\/span><\/sup>, then <\/span>determine <\/span>the <\/span>density of <\/span>gold.<\/span><\/span><\/p>\n

A. 1.93 gr\/cm3<\/sup><\/span><\/p>\n

B. 8.65 gr\/cm3<\/sup><\/span><\/p>\n

C. 19.3 gr\/cm3<\/sup><\/span><\/p>\n

D. 193 gr\/cm3<\/sup><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of gold (mgold<\/sub>) = 193 gram = 0.193 kg<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Buoyant force (FA<\/sub>) = 8000 dyne = 8 x 103<\/sup> dyne = (8 x 103<\/sup>)(10-5 <\/sup>N) = 8 x 10-2 <\/sup>N = 0.08 Newton <\/span><\/p>\n

Density of kerosene (\u03c1) = 0.8 gr\/cm3 <\/sup>= 800 kg\/m3 <\/sup><\/span><\/p>\n

Wanted :<\/u> Density of gold<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Weight of gold in air :<\/span><\/p>\n

w = m g = \u03c1b<\/sub> V g —– Equation 1<\/span><\/p>\n

V is volume of gold in kerosene.<\/i><\/span><\/p>\n

Buoyant force (FB<\/sub>) equal to the weight of gold in air (w) minus weight of gold in kerosene (w\u2019) :<\/span><\/p>\n

w \u2013 w\u2019 = FA<\/sub><\/span><\/p>\n

w \u2013 w\u2019 = \u03c1f<\/sub> V g —– Equation 2<\/span><\/p>\n

V is the volume of gold in kerosene.<\/i><\/span><\/p>\n

Both equations above can be written again below :<\/span><\/p>\n

\"Buoyant<\/span><\/p>\n

\"Buoyant<\/span><\/p>\n

The correct answer is C.<\/span><\/p>\n

    \n
  1. What is the fundamental principle behind the buoyant force?<\/strong><\/span>\n