# Buoyant force of children

• bearhug
In summary, the conversation is discussing a problem involving buoyancy and determining the number of logs needed to keep three children afloat on a log raft made of logs with a diameter of 0.30 m and a length of 1.80 m. The children each have a weight of 356 N and the density of the logs is given as 800 kg/m^3. The conversation includes discussions on finding the weight and volume of the logs, using equations for buoyancy, and calculating the effective density of the log/children combination. The final step is to find the number of logs needed to equal the weight of the displaced water.

#### bearhug

Three children, each of weight 356 N, make a log raft by lashing together logs of diameter 0.30 m and length 1.80 m. How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be 800 kg/m^3.

I'm interpreting this problem as a buoyancy problem for starters.
I have:
B=1068 N
V(obj)= 0.127
V(liq)= ? is this the same as the obj?
p(obj)= 800kg/m^3
p(liq)= 1.0e3 kg/m^3

I'm thinking I need to find the density of one log based on this information and divide the given density of all the logs by this. However I'm having a hard time finding a way to get to this using the equations for buoyancy. For a floating object the equation I was originally going to use was
p(liq)gV(liq)=p(obj)gV(obj) but I'm not sure what if the Volume of the liquid is the same as for the object since V is the volume of the fluid displaced from the object. Any guidance on this problem is greatly appreciated.

If the object is completely submerged, then the volume displaced is equal to the volume of the object.

You're given the density of the logs in the question statement.

I think you will do better thinking of the children and all logs as the floating object, so its weight is the combined weight of children and logs. Assuming the children wish to stay dry, the volume of displaced water will be no more than the volume of the logs.

Okay so I took Dan's advice and added up children and logs to get a total force. What I did was I used D=M/V and solved for M to get the mass of the logs being 101.6kg. Then used F=mg to get the force which is 995.7N and added this to (356)(3)=1068N to get 2063.7N. I used the equation F= p(obj)gV(obj) and solved for p(obj) is this right?

bearhug said:
Okay so I took Dan's advice and added up children and logs to get a total force. What I did was I used D=M/V and solved for M to get the mass of the logs being 101.6kg. Then used F=mg to get the force which is 995.7N and added this to (356)(3)=1068N to get 2063.7N. I used the equation F= p(obj)gV(obj) and solved for p(obj) is this right?
You are on the right track finding the weight of a log. The children can use as many logs as they need. You can calculate the fractional number of logs needed and then round up to get the answer. Think in terms of n logs instead of just one. You want the weight of n logs plus the weight of the children to equal the weight of water having the same volume as n logs. That way, only the logs will get wet.

You can think of the log/children combination as having an "effective density" by assuming the children have weight but no volume; their weight gets added to the weight of the logs, but the volume is just the volume of the logs. In order to float, the usual condition needs to be satisfied, which is: The _________ of the object must be ___________ the ___________ of water.

The equation in my book is V(liq)/V(obj)=p(obj)/p(liq) and by the way I'm reading your explanation I feel like it's a proportion problem. However I'm not sure how to fit weight into a similar equation, unless there's something I'm forgetting. So the weight of n logs plus children is 995.7n+1068N which has to equal the weight of the water in order to stay afloat. The volume of water is equal to 0.127n (from pie (r)^2*h). How do I use all of this to solve for the number of logs? It makes sense step by step but I just can't seem to put it all together.

bearhug said:
The equation in my book is V(liq)/V(obj)=p(obj)/p(liq) and by the way I'm reading your explanation I feel like it's a proportion problem. However I'm not sure how to fit weight into a similar equation, unless there's something I'm forgetting. So the weight of n logs plus children is 995.7n+1068N which has to equal the weight of the water in order to stay afloat. The volume of water is equal to 0.127n (from pie (r)^2*h). How do I use all of this to solve for the number of logs? It makes sense step by step but I just can't seem to put it all together.
You can find the weight of the water from its volume and density. n has to be big enough so that the weight of displaced water is at least as much as the weight of the n logs plus children.

bearhug said:
Three children, each of weight 356 N, make a log raft by lashing together logs of diameter 0.30 m and length 1.80 m. How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be 800 kg/m^3.

I'm interpreting this problem as a buoyancy problem for starters.
I have:
B=1068 N
V(obj)= 0.127
V(liq)= ? is this the same as the obj?
p(obj)= 800kg/m^3
p(liq)= 1.0e3 kg/m^3

I'm thinking I need to find the density of one log based on this information and divide the given density of all the logs by this. However I'm having a hard time finding a way to get to this using the equations for buoyancy. For a floating object the equation I was originally going to use was
p(liq)gV(liq)=p(obj)gV(obj) but I'm not sure what if the Volume of the liquid is the same as for the object since V is the volume of the fluid displaced from the object. Any guidance on this problem is greatly appreciated.

how do you know this: V(obj)= 0.127 ? and is this the volume of the logs or of the children?

## 1. What is the buoyant force of children?

The buoyant force of children refers to the upward force that acts on their bodies when they are submerged in a fluid, such as water. This force is equal to the weight of the fluid that is displaced by the child's body.

## 2. How is the buoyant force of children calculated?

The buoyant force of children is calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by an object. To calculate this force, the weight of the child and the density of the fluid must be known.

## 3. Does the buoyant force of children vary based on their weight?

Yes, the buoyant force of children will vary based on their weight. The heavier the child, the greater the buoyant force they will experience when submerged in a fluid.

## 4. How does the buoyant force of children compare to that of adults?

The buoyant force of children is generally greater than that of adults, as children have a higher ratio of surface area to weight compared to adults. This means that more of their body is in contact with the fluid, resulting in a greater buoyant force.

## 5. Can the buoyant force of children be affected by their body composition?

Yes, the buoyant force of children can be affected by their body composition. Children with a higher percentage of body fat will experience a greater buoyant force compared to children with a lower percentage of body fat, as fat is less dense than muscle and bone.