What is the correct equation to use to calculate the resultant force for the below inclined plane Freebody diagram?

If you and your friend are riding your bikes side by side and your reference points are each other, are you in motion? Explain y

geniusboy [140]

If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary

6 0

3 years ago

A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a

grigory [225]

Answer:

a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

-w x - W x₂ + R y = 0 (1)

usemso trigonometry to find distances

cos 60.08 = x / 7.5

x = 7.5 cos 60.08

x = 3.74 m

fireman

cos 60.08 = x₂ / 4

x2 = 4 cos 60

x2 = 2 m

wall support

sin 60.08 = y / 15

y = 15 are 60.08

y = 13 m

we substitute in equation 1

R y = w x + W x2

R = (w x + W x2) / y

R = (500 3.74 +800 2) / 13

R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

R -fr = 0

R = fr

fr = 266.92 N

Y Axis

Fy - w-W = 0

fy = 500 + 800

fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

cos 60.08 = x₃ / 9.00

x₃ = 9 cos 60

x₃ = 5.28 m

from equation 1

R = (w x + W x₃) / y

R = 500 3.74 + 800 5.28) / 13

R = 468.769 N

we saw that

fr = R = 468.769

The expression for the friction force is

fr = μ N

in this case the normal is the ratio to pesos

N = Fy

N = 1300 N

μ N = fr

μ = fr / N

μ = 468,769 / 1300

μ = 0.36

7 0

3 years ago

What is the equivalent resistance of the

BigorU [14]

Answer:

Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

Since $R_{1}$ and $R_{2}$ are connected in parallel, the voltage across the two resistors would both be $V$ . Thus, the current going through the two resistors would be $(V / R_{1})$ and $(V / R_{2})$ , respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .