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d1i1m1o1n [39]
4 years ago
9

The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe

en the two beams and between the lower beam and the bed of the trailer are 0.25 and 0.30, respectively. Knowing that the load does not shift, determine(a) the maximum acceleration of the trailer and the corresponding tension in the cable, (b) the maximum deceleration of the trailer.
Physics
1 answer:
Novosadov [1.4K]4 years ago
4 0

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

          a = (2μ₁ + μ₂) g

          a = (2 0.30 + 0.25) 9.8

          a= 8.33 m/s²

Let's look for cable tension with beam 2

          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

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A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia
Ierofanga [76]

Answer:

v_{0}=319.2 m/s    

Explanation:

We need to use the momentum and energy conservation.

p_{0}}=p_{f}

mv_{0}=(m+M)V_{1}

Where:

  • m is the mass of bullet (m=0.01 kg)
  • M is the mass of the wooden (M=0.75 kg)
  • v(0) initial velocity of bullet
  • V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.            

(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh

Here:

  • V(1) is the velocity of the combined at the initial position
  • h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

cos(\theta) =(L-h)/L=(1.72-0.8)/1.72

\theta =57.66

Now, we can solve the equation to find V(2)

V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}

V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}

V_{2}=1.40 m/s        

Now we can find V(1) using the conservation of energy equation

(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh

V_{1}=\sqrt{V_{2}^{2}+2gh}

V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}          

V_{1}=4.20 m/s        

Finally, using the momentum equation we find v(0)

v_{0}=\frac{(m+M)V_{1}}{m}                

v_{0}=\frac{(0.01+0.75)*4.20}{0.01}

v_{0}=319.2 m/s        

I hope it helps you!

 

7 0
4 years ago
A 23kg block at rest on a flat surface experiences 200 N of friction. What is the coefficient of friction for the surface
Brrunno [24]

Ff=μN=μmg=>μ=Ff/mg=200/230=0.86

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Answer:

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The distance traveled is clearly 80km since it's all done in the same direction, we only need to know the time elapsed. For this we calculate the time elapsed on the first part, and add it to the time elapsed on the second part using always the formula \Delta t=\frac{\Delta x }{v}, where v is the velocity on each part, which is constant.

The time elapsed for the first part is \Delta t_1=\frac{40 km}{30km/h}=\frac{4}{3}h, and the time elapsed for the second part is \Delta t_2=\frac{40 km}{60km/h}=\frac{2}{3}h, giving us a total time of \Delta t_1+\Delta t_2=\frac{4}{3}h+\frac{2}{3}h=2h.

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