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Lemur [1.5K]
3 years ago
15

A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, r

espectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Physics
1 answer:
V125BC [204]3 years ago
7 0

Answer:

(a) whether slipping occurs between the belt and the cylinder i think is the answer dont hate if you get it wrong please and thank you.

Explanation:

i am just guessing otay.

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4 0
3 years ago
You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
iren [92.7K]

Answer:

angle minimum   θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

     fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

     

the friction force formula is

     fr = μ N

     fr = μ W

we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

      μ sin θ - cos θ + ½ cos θ= 0

         

       μ sin θ - ½ cos θ = 0

       sin θ / cos θ = 1/2 μ

       tan θ = 1/2 μ

       θ = tan⁻¹ (1 / 2μ)

       θ = tan⁻¹ (1 (2 0.57))

      θ = 41.3º

7 0
3 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

6 0
4 years ago
Please help it’s do soon
viktelen [127]
When y’all are all over the place and I wanna is a time to come sit in my
8 0
3 years ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0
3 years ago
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