Physics, 03.03.2020 02:54 randall10

If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.

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Answer from: Nobleufia

The force developed in the link is 6.36 N.

Explanation:

Given that,

Mass of block A = 10 kg

Mass of block B = 6 kg

Coefficients of kinetic friction $\mu_{A}= 0.1$

Coefficients of kinetic friction $\mu_{B}= 0.3$

Suppose the angle is 30°

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}$

Put the value into the formula

$a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}$

$a=3.415\ m/s^2$

We need to calculate the force developed in the link

For block A,

Using balance equation

$ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T$

$T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta$

Put the value into the formula

$T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30$

$T=-6.36\ N$

Negative sign shows the opposite direction of the force.

Hence, The force developed in the link is 6.36 N.

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Answer from: Quest

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If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released...

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