All categories

2 years ago

Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.

2 answers:

6 0

CASE 1: Direction vertically downward on the blade

Assume pin force at E acts downward on the blade and CW is a positive moment 
Ey[3.4] - F[1.7] = 0 
Ey = F/2 

Case 2: Direction vertically upward on the blade . Assume CW is positive moment 

Ey[1.5sin19] – P[21 – 1.5sin19] = 0 
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

Alexus [3.1K]2 years ago

5 0

If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0
Ey = F/2
and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

You might be interested in

A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en

nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy $E_{n}$ of a particle of mass m in the nth energy state of an infinite square well potential with width L is:

$E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

$E_{1}=\frac{h^{2}}{8mL^{2}}$

$E_{2}=\frac{h^{2}}{2mL^{2}}$

So we can rewrite the energy in the ground state as:

$E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

$E_{1}=\frac{1}{4} E_{2}$

$E_{1}=\frac{1}{4} ( 6.0\ eV)$

Finally

$E_{1}=1.5\ eV$

6 0

2 years ago

A spring stores 10. joules of elastic potential

Mekhanik [1.2K]

The answer would be . Since we are looking for the spring constant you would need to use the formula
$PEs = \frac{1}{2} k {x}^{2}$
. Then you'd substitute, for PEs and x.
$10j=1/2k(.20m^2).$
Then solve. k=500n/m

6 0

2 years ago

Read 2 more answers

Arwan finds a piece of quartz while hiking in the mountains. When he returns to school, he takes it to his science teacher. She

bekas [8.4K]

it's 303.4 cm3. i just took the test

8 0

2 years ago

Read 2 more answers

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N

Mnenie [13.5K]

Answer:

8N and 32N

Explanation:

Given that a light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.

To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.

Let the two upward forces produce by the saw horses be P1 and P2

Assuming that the weight is negligible

Sum of the upward forces = sum of the downward forces.

P1 + P2 = 40 ....... (1)

Also, the sum of the clockwise moment = sum of the anticlockwise moments.

Let's assume that the board is uniform. The weight will act at the centre.

Taking moment at the centre:

P1 × 5 + 40 × 2 = 0

P1 = 40 / 5

P1 = 8N

Substitute P1 into equation 1

8 + P2 = 40

P2 = 40 - 8

P2 = 32N

3 0

2 years ago

Read 2 more answers