How can the motion of an object that is NOT moving change?

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

What kind of habitat do we live in

Aleksandr-060686 [28]

the worst because its not enough habitat and when the cutting down a whole lot of rainforest at a time which is killing animals escpecialy rainforest animals

8 0

3 years ago

what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can

Anon25 [30]

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

8 0

3 years ago

A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver

MissTica

Answer:

Approximately $2.31\; \rm m \cdot s^{-2}$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assuming that $g = 9.81\; \rm m \cdot s^{-2}$ the weight on this 72-kg skydiver would be $W = m \cdot g = 72 \; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 706.32\; \rm N$ (points downwards.)

Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.

Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:

$\begin{aligned}F(\text{net force}) &= W - F(\text{air resistance})\\ &= 706.32\; \rm N - 540\; \rm N =166.32\; \rm N \end{aligned}$ .