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A person is pushing two carts that are connected with a metal bar so that the carts are moving at constant acceleration 0.30m/s2. The masses of the carts are 100 kg and 150kg, and the mass of the connecting bar is 60 kg. Assume that the friction forces are negligible. Also assume that you are pushing the lighter cart to the right. A) Determine the force that the bar exerts on the lighter cart. B) Determine the force that the heavier cart exerts on the bar.

Respuesta :

Answer:

63N, 45N

Explanation:

Given:

Tension 1 = T1

We have the following free body equations:

F - T1 = 100*0.3 -(I)

T1 - T2 = 60*0.3 -(II)

T2 = 150*0.3 -(III)

Adding the three above equations, we have

F - T1 + T1 - T2 + T2 = (100*0.3) + (60*0.3) + (150*0.3)

Hence, F = 93N

Hence, substituting into equation I we have

T1 = Force by bar on lighter cart = 93 - (100*0.3) = 63N

Hence, force that heavier cart exerts on bar

T2 = 150*0.3 = 45N

a. The force that the bar exerts on the lighter cart should be 63 N.

b. The force that the heavier cart exerts on the bar is 45 N

Calculation of force:

(A)

Here we considered the following free body equations:

F - T1 = 100*0.3 -(I)

T1 - T2 = 60*0.3 -(II)

T2 = 150*0.3 -(III)

Now we have to add these equation

So,

= F - T1 + T1 - T2 + T2

= (100*0.3) + (60*0.3) + (150*0.3)

F = 93N

Now

T1 = Force by bar on lighter cart = 93 - (100*0.3) = 63N

(B)

Now

The force that heavier cart exerts on bar

T2 = 150*0.3

= 45N

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