Need help finding the buoyant force in this problem

[TheKShaugh]

Hi everyone,

This question is killing me, I don't know why but I just can't get my head around it.

Homework Statement

A team of 6 scouts plans to cross a lake on a raft they designed.

The scouts have wooden beams with an average density of .80. The beams measure 30cm x 30cm x 3m.

The average mass for the team members is 65kg, and for their safety, they want the top of the raft to be at least 3 cm above the surface of the water.

How many beams will they need to complete their project successfully?

Homework Equations

I'm not sure if I need to post them. f = mg is the main one.

The Attempt at a Solution

My tutor gave me this method:

Fg(scouts) = 65 x 9.8 x 6

V(beam) = .27m^3

Fg(beam) = 800 x 9.8 x .27

Fg(raft) = Fg(beam) x n [n = number of beams]

Fg(total) = Fg(scouts) + Fg(raft)

Fb = 2381 x n

Fb = Fg(total) = Fg(scouts) + Fg(raft).

Solve for n.

Now I'm just a lowly high school student so my math isn't all that great, and when I do this I end up getting 0. The equation as I write it out is

2381n = -3822 - 2116n = -3822 - 2116n

6713n = -3822 +3822

6713n = 0

Could anyone help me out with this question?

Thanks,

Kevin

 

[SteamKing]

Your algebra has taken a wrong turn.

Fg = -3822 - 2116n

Fb = 2381n

From statics, we know that Fb + Fg = 0 (there is no net force on a floating object)

so, 2381n -3822 - 2116n = 0

Solve for n

 

[haruspex]

2381n = -3822 - 2116n = -3822 - 2116n

6713n = -3822 +3822

Try that step again.

 

[TheKShaugh]

Your algebra has taken a wrong turn.

Fg = -3822 - 2116n

Fb = 2381n

From statics, we know that Fb + Fg = 0 (there is no net force on a floating object)

so, 2381n -3822 - 2116n = 0

Solve for n

Thank you. You truly are the king of kings.

 

[Nickie]

Hello Kevin,

I can understand why this problem may be confusing for you. Let's break it down step by step.

First, we need to understand the concept of buoyant force. This is the force exerted by a fluid (in this case, water) on an object immersed in it. It acts in the opposite direction of gravity and is equal to the weight of the displaced fluid.

In this problem, the scouts want the top of their raft to be at least 3 cm above the surface of the water. This means that the buoyant force must be equal to the weight of the raft plus the weight of the scouts.

To find the weight of the raft, we need to know its volume and density. The volume of the raft is 30cm x 30cm x 3m = 0.27m^3. The density of the wooden beams is given as 0.80, so the weight of the raft is 0.80 x 0.27 x 9.8 = 2.106 N.

Next, we need to find the weight of the scouts. This is given as 65 kg x 9.8 = 637 N.

Therefore, the total weight that the buoyant force must support is 2.106 N + 637 N = 639.106 N.

Now, we can use the formula Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid (in this case, water), V is the volume of the displaced fluid, and g is the acceleration due to gravity (9.8 m/s^2).

We know that the scouts want the top of the raft to be at least 3 cm above the surface of the water, so the volume of the displaced fluid is 0.27m^3 + 0.03m^3 = 0.3m^3.

Plugging in the values, we get:

Fb = 1000 x 0.3 x 9.8 = 2940 N

Since we want the buoyant force to be equal to the total weight of the raft and scouts, we can set up the equation:

2940 N = 639.106 N + 0.8 x 0.27 x n x 9.8

Solving for n, we get n = 22. So, the scouts will need 22