Determining maximum force applied on a lever

1. Jul 2, 2009

KEØM

1. The problem statement, all variables and given/known data
The lever BCD is hinged at C and is attached to a control rod at B. Determine the maximum force $$\vec{P}$$ which can be safely applied at D
if the maximum allowable value of the reaction at C is 500 N.

Here is a picture of the problem. It is number 4.20.
"[URL [Broken] title="download file from Jumala Files"]http://jumalafiles.info/showfile2-14173678132920785343298714637103960/problem420.pdf [Broken][/URL]

2. Relevant equations

$$\Sigma F_{x} = 0$$

$$\Sigma F_{y} = 0$$

$$\Sigma M_{C} = 0$$

3. The attempt at a solution

I know I have two unknowns at pin C $$C_{x}, C_{y}$$ but I don't know either of their magnitudes. I have a tension in rod BA and I don't its magnitude and I also don't know the magnitude of $$\vec{P}.$$ I only have 3 equations to work with here but more than that of unknowns. Am I missing a relationship here or am I not making a correct assumption?

KEØM

Last edited by a moderator: May 4, 2017
2. Jul 3, 2009

nvn

But you also have a given equation regarding the magnitude of C. Therefore, don't you have four equations?

3. Jul 3, 2009

KEØM

I must just not be seeing how that equation helps me because I can't solve for any thing right now.

$$\Sigma M_{B} = -(.03)C_{x} - (.04)C_{y} - (.105)P = 0$$

$$\Sigma F_{y} = C_{y} + Tsin(\theta) = 0$$

$$\Sigma F_{x} = C_{x} + P + Tcos(\theta) = 0$$

$$C = 500 = \sqrt{(C_{x})^2 + (C_{y})^2}$$

$$\theta = 53.1^{\circ}$$

4. Jul 3, 2009

nvn

Last edited by a moderator: Apr 24, 2017
5. Jul 4, 2009

KEØM

I have tried several times in solving these four equations and I just can't do it. Are you sure I am not missing some assumption or relationship?

6. Jul 4, 2009

nvn

Your equations are correct. Use the advice in post 4. If you show your work, it will quickly uncover your algebra mistake.

7. Jul 4, 2009

KEØM

$$\Sigma M_{B} = -(.03)C_{x} - (.04)C_{y} - (.105)P = 0$$

$$\Sigma F_{y} = C_{y} + Tsin(\theta) = 0$$

$$\Sigma F_{x} = C_{x} + P + Tcos(\theta) = 0$$

$$C = 500 = \sqrt{(C_{x})^2 + (C_{y})^2}$$

$$\theta = 53.1^{\circ}$$

$$C_{y} + Tsin(\theta) = 0 \Rightarrow C_{y} = -\frac{4}{5}T$$

$$C_{x} + P + Tcos(\theta) = 0 \Rightarrow C_{x} + P + \frac{3}{5}T = 0$$

$$T = -\frac{5}{4}C_{y}$$

$$C_{x} -\frac{1}{4}C_{y} = -P$$

$$-(.03)C_{x} - (.04)C_{y} - (.105)P = 0 \Rightarrow .2857C_{x} + 0.3810C_{y} = -P$$

$$C_{x} -\frac{1}{4}C_{y} = .2857C_{x} + 0.3810C_{y}$$

$$C_{x} -\frac{1}{4}C_{y} = 2857C_{x} + 0.3810C_{y}$$

$$0.7143C_{x} = 0.6310C_{y} \Rightarrow 1.13C_{x} = C_{y}$$

$$500^2 = C_{x}^2 + (1.13C_{x})^2$$

$$C_{x} = 331.02 N, C_{y} = 374.74 N$$

While I was typing my failed attempt in here I finally realized how to solve it.

Is this right? When I try to solve for P I am getting a negative answer.

Thanks again for helping me nvn,

KEØM

8. Jul 4, 2009

nvn

Close, but not quite; (3/5)*(5/4) is not equal to 1/4. Try it again. Also, always place a zero before the decimal point for numbers less than 1.

9. Jul 4, 2009

KEØM

Woops! Stupid Mistakes. Now I get:

$$C_{x} = 422.7 N, C_{y} = 267.0 N$$

This doesn't change the negative value I get for P though.

10. Jul 4, 2009

nvn

Very good, except notice that when you solve for Cx, it is actually Cx = +/-(number)^0.5. Knowing P is positive, you can figure out whether the other +/- values should be positive or negative.

11. Jul 4, 2009

KEØM

Ok. So if I choose $$C_{x}$$ to be negative then:

$$\frac{3}{4}C_{y} - C_{x} = P \Rightarrow \frac{3}{4}(267.0) - (-422.7) = P$$

$$P = 623.0 N$$ and I have my answer

Thanks again,

KEØM

12. Jul 4, 2009

nvn

No, plug your answer for Cx into an earlier equation to solve for Cy, to obtain the correct sign on Cy.