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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a mass of m2 = 86 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 and the coefficient of kinetic friction between the two crates is μk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?
2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?
3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?
4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s2
5)As the upper crate slides, what is the acceleration of the lower crate?

Respuesta :

Chenk13

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

1) a = 2.13 m/s²

2) F = 51.12 N

3) T_max = 851.62 N

4)a₁ = 6.08 m/s²

5) a₂ = 12.11 m/s²

This is based on the concept of frictional motion with coefficients of friction.

We are given;

Mass of smaller top crate; m₁ = 24 kg

Mass of larger bottom crate; m₂ = 86 kg

Coefficient of static friction; μ_s = 0.79

Coefficient of kinetic friction; μ_k = 0.62

For the top crate, the sum of forces will be expressed as;

F = m₁a₁  - - - (eq 1)

For the bottom crate, the sum of forces will be expressed as;

T - F = m₂a₂  - - - (eq 2)

Where F is frictional force exerted by small crate on big crate and T is tension.

1) We are told that the rope is pulled with a tension T = 234 N. And that the top crate will not slide. Thus, the acceleration of the small crate will be gotten from a combination of equation 1 and 2 to get the formula;

T = m₂a + m₁a

Making a which is the acceleration the subject gives;

a = T/(m₁+m₂)

Plugging in the relevant values gives;

a = 234/(24 + 86)

a = 2.13 m/s²

2) From eq(1) above, we see that the formula for the frictional force the small crate exerts on the big one is;

F = m₁a₁

Thus, Plugging in the relevant values gives;

F = 24 × 2.12

F = 51.12 N

3) maximum tension that the lower crate can be pulled at before the upper crate begins to slide will occur at the maximum friction. The formula for mx friction is;

(F₂)_max = μ_s•m₂•g

Using the concept of eq(1) We can say that;

F₂ = m₂a

m₂a = μ_s•m₂•g

m₂ will cancel out to give;

a = μ_s • g

From first answer earlier, we saw that;

T = m₂a + m₁a

T = (m₂ + m₁)a

Plugging in the relevant values gives;

T_max = (m₁ + m₂)μ_s*g

T_max = (24 + 86) × 0.79 × 9.8

T_max = 851.62 N

4) To get the acceleration of the upper crate, we will make use of the formula for kinetic friction which is:

F = μ_k*m₁g

From earlier, we saw that; F = m₁a₁

Thus;

m₁a₁ = μ_k*m₁g

m₁ will cancel out to get;

a₁ = μ_k*g

a₁ = 0.62 × 9.8

a₁ = 6.08 m/s²

5) the acceleration of the lower crate as the upper crate slides will be gotten by putting μ_k*m₁*g for m₁a₁ in eq(2) to get;

m₂a₂ = T - (μ_k*m₁*g)

Making a₂ the subject gives us;

a₂ = (T - μ_k•m₁•g)/m₂

Plugging in the relevant values;

a₂ = (1187 - (0.62 × 24 × 9.8))/86

a₂ = 12.11 m/s²

Read more at; brainly.com/question/4853862

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