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gtnhenbr [62]
3 years ago
10

A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ​∘ ​​ incline with constant speed under the action o

f a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?
Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

11.82 N upward and parallel to the incline.

Explanation:

Let g = 9.81 m/s2. Gravity that acts on the bin has a magnitude of

W = mg = 3*9.81 = 29.43 N

The gravity component that is parallel to the 40 degree incline is

Wsin40^0 = 29.43*0.643 = 18.9 N

If the bin is moving at constant speed, then the net force (parallel to the incline) is also 0. Gravity is balanced with the push force of 26 N and the friction force.

The magnitude of friction force is 26 - 18.9 = 7.08 N

The the bin is moving down the incline, then gravity is 18.9N , friction is 7.08 N, one must create an upward force of 18.9 - 7.08 = 11.82 N parallel with the incline to balance this.

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Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless  and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of  the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting  on both masses, we can write both equations as follows:

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Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the  horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

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Answer:

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