Finding the forces in equalibrium

Ester

A woman who weighs 476 N is leaning against a smooth vertical wall at an agnle 60 degrees above the horizontal. Find (a) the force FN (directed perpendicular to the wall) exerted on her shoulders by the wall and the (b) horizontal and (c) vertical components of the force exerted on her shoes by the ground.

I know that the the vertical component is 476 N, the other one's I have no clue of how to do. I made the sum of the forces in the x direction zero. So I know that the force of static friction will equall the normal force exerted by the wall. I also made the sum of the torques to equal zero. So, FnLn = FwLw. I get my answer to be 137.41 N, but this is not right. Ln represents the lever arm's lengh. I used trig functions to get Lw=cos60 (0.75), and Ln=sin60 (1.5)

This is the drawing they provide the distance (hypotenuse) from the Fn to the ground as 1.5. This number is broken into 1.1 m for her feet to the waist and 0.4 m from waist to shoulders. The head isn't included since there is no force for it.

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Ester

It's a tricky question, I don't understand what's going on and how to solve it. Please help me if you can. :shy:

Last edited:

Doc Al

Mentor
location of center of mass?

I suspect you are meant to assume that the woman's center of mass is at her waist, not midway between her feet and shoulders (as you assumed in your caculation).

Ester

yes that is true, but still, how would you solve it?

OlderDan

Homework Helper
Ester said:
yes that is true, but still, how would you solve it?
Use the corrected distance for Lw. You know Fw and Ln, so you can solve for Fn, and, as you have stated, that must equal the magnitude of horizontal force of static friction exerted on her shoes by the ground.

amt

Always post a free body diagram first. This makes things easy for both yourself and the rest of us trying to help.

Wall
|\
|.\
|..\
|...\
|....\
|.....\
|......\
|.......\
|........\
|......60\
---------

From this diagram how can you conclude that the vertical component is her weight? The length of the hypotonuse is 476N. Whenever you see angles like this, you need to break it down into 'x' and 'y' components. Let's call that Fx and Fy. Here is a clue- Fx is the length of the base of the triangle and Fy is the height of the triangle. Remember every force has an equal and opposite force. Now in order to find the length of the base and height of the triangle, can you write out the trig equations?

Look at this example (problem #2) it's quite similar:

Please post a free body diagram of your understanding.

Doc Al

Mentor
Ester said:
yes that is true, but still, how would you solve it?
As OlderDan points out, you'd solve it just like you did in your first post. Only this time using the correct location of the woman's weight (the corrected distance Lw). The method you were using looked OK to me.

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