# 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.

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

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.