If the mass of block A = 10,kg,the coefficien | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To start the motion,the tension $T$ in the string must overcome the limiting static friction force $f_{s,max}$.For block $A$ on the horizontal sur

Vedclass · Accepted Answer

$2\,kg$

Vedclass · Answer

(A) To start the motion,the tension $T$ in the string must overcome the limiting static friction force $f_{s,max}$.For block $A$ on the horizontal surface,the normal force $N = m_A g$.The limiting static friction is $f_{s,max} = \mu_s N = \mu_s m_A g$.For block $B$ hanging vertically,the tension $T = m_B g$.At the point of starting motion,the system is in limiting equilibrium,so $T = f_{s,max}$.Substituting the values: $m_B g = \mu_s m_A g$.$m_B = \mu_s m_A = 0.2 	imes 10\,kg = 2\,kg$.Therefore,the mass of block $B$ required to start the motion is $2\,kg$.

If the mass of block $A = 10\,kg$,the coefficient of static friction $\mu_s = 0.2$,and the coefficient of kinetic friction $\mu_k = 0.2$,find the mass of block $B$ required to start the motion.

Similar Questions

Explain: "Static friction force opposes impending motion".

The maximum value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,N$)

The limiting friction between two bodies in contact is independent of

$A$ body of mass $5 \,kg$ is placed on a rough horizontal surface of coefficient of static friction $\frac{1}{3}$. The least pulling force to be applied on the body at an angle $45^{\circ}$ with the horizontal to slide it,is . . . . . . $\left(g=10 \,ms^{-2}\right)$

For a given mass,how does rolling friction compare to static friction and kinetic friction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module