Question

A 100-lb force directed vertically downward is applied to the toggle vise at C. Knowing that link BD is 6 in. long and that a=4 in., determine the horizontal force exerted on block E.

Answer 1

Formulate an expression for the surface energies of ideal (100),(110), and (111) surfaces of a material that adopts a simple cubic crystal structure. Leave your expression for γ in terms of a and ΔHsub. ​.

the calculation step by step.

Given:

1. F_C: The force applied vertically downward at point C = 100 lbs

2. BD: The length of link BD = 6 inches

3. a: The horizontal distance from A to the line of action of force F_C = 4 inches

4. θ: The angle that link AB makes with the horizontal = 15° (this will be used to calculate the horizontal distance that force F_C acts from point B)

Here is how we would set up the problem:

Step 1: Identify the System and Assumptions

- The system is in static equilibrium, which means the sum of the forces and moments in any direction equals zero.

- We assume all joints are pin joints, and the weight of the links can be neglected.

Step 2: Define the Equilibrium Equations

For static equilibrium of a body, the sum of the moments about any point is zero. We will calculate moments about point B because it eliminates the need to consider forces at point B (since their moment arm would be zero).

Step 3: Calculate the Moment Arm for F_C

Since AB is at an angle of 15°, we will calculate the horizontal distance from B to where F_C is applied using the length of BD and the angle θ:

`\[ \text{Horizontal distance} = a + (\text{BD} * \cos(θ)) \]`

Now, plugging in the numbers:

`\[ a = 4 \text{ inches}`

`\[ \text{BD} = 6 \text{ inches} \]`

`\[ θ = 15° \]`

`\[ \cos(θ) = \cos(15°) \]`

Step 4: Moment Equation about Point B

Set up the moment equation about point B where clockwise moments are set to equal counter-clockwise moments. The force at C creates a clockwise moment, and the horizontal force exerted on E creates a counter-clockwise moment.

`\[ F_C * (\text{Horizontal distance}) = F_E * (\text{BD}) \]`

Step 5: Solve for F_E

Now, we will substitute the values and solve for F_E:

`\[ 100 * (4 + 6 * \cos(15°)) = F_E * 6 \]`

Step 6: Calculate F_E

`\[ F_E = (100 * (4 + 6 * \cos(15°)))/(6) \]`

Step 7: Numerical Substitution and Computation

Now let's do the numerical substitution and computation:

`\[ \cos(15°) \approx 0.9659 \]`

S

`\[ F_E = (100 * (4 + 6 * 0.9659))/(6) \]`

`\[ F_E \approx (100 * (4 + 5.7954))/(6) \]`

`\[ F_E \approx (100 * 9.7954)/(6) \]`

`\[ F_E \approx (979.54)/(6) \]`

`\[ F_E \approx 163.26 \text{ lbs} \]`

Thus, the horizontal force exerted on block E is approximately 163.26 lbs.

The diagram of the question given below: