Answer:
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 [tex]\mu_{A}= 0.1[/tex]
Coefficients of kinetic friction [tex]\mu_{B}= 0.3[/tex]
Suppose the angle is 30°
We need to calculate the acceleration
Using formula of acceleration
[tex]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}}[/tex]
Put the value into the formula
[tex]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}[/tex]
[tex]a=3.415\ m/s^2[/tex]
We need to calculate the force developed in the link
For block A,
Using balance equation
[tex]ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T[/tex]
[tex]T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta[/tex]
Put the value into the formula
[tex]T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30[/tex]
[tex]T=-6.36\ N[/tex]
Negative sign shows the opposite direction of the force.
Hence, The force developed in the link is 6.36 N.
