Answer:
Mass of object B is 9 kg for which block A remains at rest.
Explanation:
Given:
Mass of the block A, [tex]m_1[/tex] = 10 kg
Co-efficient of friction, [tex]\mu _s[/tex] = 0.4
Angle of inclination, [tex]\theta[/tex] = 35°
Mass of block B is unknown.
Have a look to the FBD of the inclined surface along with the two masses and the friction forces and the component of downward weight W (i.e mg).
Let the mass of the block B is [tex]m_2[/tex] .
Forces acting on [tex]m_1[/tex] .
⇒ Net horizontal force, [tex]T-(m_1gcos\theta + \mu_sN)=0[/tex] ...(i)
⇒ Net vertical force, [tex]N=m_1gcos(\theta)[/tex] ...(ii)
Plugging value of N from equation (ii) to (i).
T (the force on the string) can be written as.
⇒ [tex]T=m_1gsin(\theta)+\mu_sm_1gcos(\theta)[/tex]
⇒ [tex]T=m_1g(sin\theta+\mu_scos\theta)[/tex] ...equation (iii)
Net forces acting on [tex]m_2[/tex] .
⇒ [tex]T-m_2g=0[/tex]
⇒ [tex]T=m_2g[/tex] ...equation (iv)
Equating (iv) and (iii) as both are string forces on the same pulley.
⇒ [tex]m_2g=m_1g(sin\theta+\mu_scos\theta)[/tex]
⇒ [tex]m_2=m_1(sin\theta+\mu_scos\theta)[/tex]
⇒ Plugging the values.
⇒ [tex]m_2=10[sin(35)+0.4cos(35)][/tex]
⇒ [tex]m_2=9[/tex] kg
And [tex]m_2=m_B=9[/tex] kg
So,
The largest mass of [tex]m_B[/tex], attached to the dangling end, for which [tex]m_1[/tex] remains at rest is of 9 kg.
