Question
A gymnast is attempting to perform splits. From the information given in Figure 9.35, calculate the magnitude and direction of the force exerted on each foot by the floor.
<b>Figure 9.35</b> A gymnast performs full split. The center of gravity and the various distances from it are shown.
Figure 9.35 A gymnast performs full split. The center of gravity and the various distances from it are shown.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

1.11×103 N, 18.41.11\times 10^{3}\textrm{ N, } 18.4^\circ above horizontal toward the center of gravity.

Solution video

OpenStax College Physics, Chapter 9, Problem 16 (Problems & Exercises)

OpenStax College Physics, Chapter 9, Problem 16 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .

Calculator Screenshots

  • OpenStax College Physics, Chapter 9, Problem 16 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to calculate the force on each foot of this gymnast who's doing the splits. We have to make some assumptions because there isn't really enough information given to properly answer the question; it wasn't clear to me whether the force on the foot was straight up or whether it should be along the legs but since it is asking for the direction of the force, we are gonna assume that they must be expecting some answer that's more complicated than just straight up. So we are gonna have the force going along the leg so it's at this angle Θ which we can calculate because we know that tangent Θ will be opposite this height to the center of gravity divided by this adjacent which is the length of the leg horizontally. So that's 0.30 divided by 0.90 is tan of Θ and which means that Θ is the inverse tan of that which is 18.4 degrees. So we know that the vertical components of these two forces have to balance the gravity downwards. So 2 times F times sin Θ— which gives the opposite leg of this force here— has to equal the weight downwards and we can rearrange this for F by dividing both sides by 2sin Θ and we have that F is the weight divided by 2sin Θ. So that's 700 newtons divided by 2 times sin of 18.435 degrees giving us an answer of 1.11 times 10 to the 3 newtons, 18.4 degrees above horizontal towards the center of gravity.

Comments

Hi Lauren, thank you for the question. Consider one foot: we assume a force is being applied on the foot due to the ground in a direction along the leg. This force is acting at an angle theta between the leg and the ground. This force needs to play a part in balancing gravity. Gravity is straight down, so the vertical component of this force along the leg is what balances gravity. The question then is how to calculate the component of this force along the leg which is straight up. This component is FsinθF\sin{\theta} for one foot, and since there are two feet that experience an equal force, we write that 2Fsinθ=w2F\sin{\theta} = w where ww is the weight downwards.
Hope this helps,
Shaun