In the shown arrangement,if f_A,f_B and T are | vedclass

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(A) For block $A$ (mass $m = 5\,kg$,angle $	heta_A = 37^{\circ}$):Normal force $N_A = mg \cos 37^{\circ} = 5 	imes 10 	imes 0.8 = 40\,N$.Maximum st

Vedclass · Accepted Answer

$f_A = 5\,N, f_B = 15\, N, T = 25\, N$

Vedclass · Answer

(A) For block $A$ (mass $m = 5\,kg$,angle $	heta_A = 37^{\circ}$):Normal force $N_A = mg \cos 37^{\circ} = 5 	imes 10 	imes 0.8 = 40\,N$.Maximum static friction $(f_{\max})_A = \mu N_A = 0.5 	imes 40 = 20\,N$.Component of gravity down the incline $W_A = mg \sin 37^{\circ} = 5 	imes 10 	imes 0.6 = 30\,N$.For block $B$ (mass $m = 5\,kg$,angle $	heta_B = 53^{\circ}$):Normal force $N_B = mg \cos 53^{\circ} = 5 	imes 10 	imes 0.6 = 30\,N$.Maximum static friction $(f_{\max})_B = \mu N_B = 0.5 	imes 30 = 15\,N$.Component of gravity down the incline $W_B = mg \sin 53^{\circ} = 5 	imes 10 	imes 0.8 = 40\,N$.Since $W_B > W_A$,block $B$ tends to move down the incline. The system is in equilibrium because the net driving force $(W_B - W_A = 10\,N)$ is less than the total maximum friction $(f_{\max})_A + (f_{\max})_B = 20 + 15 = 35\,N$.For block $B$: $W_B - T - f_B = 0 \Rightarrow 40 - T - f_B = 0 \Rightarrow T + f_B = 40$.For block $A$: $T - W_A - f_A = 0 \Rightarrow T - 30 - f_A = 0 \Rightarrow T = 30 + f_A$.Since the system is at rest,$f_B$ will reach its maximum value to oppose motion: $f_B = 15\,N$.Substituting $f_B = 15\,N$ into the equation for $B$: $T + 15 = 40 \Rightarrow T = 25\,N$.Now,find $f_A$ using $T = 30 + f_A$: $25 = 30 + f_A \Rightarrow f_A = -5\,N$. The negative sign indicates friction acts in the opposite direction to oppose the tendency of motion. Thus,the magnitude is $5\,N$.

In the shown arrangement,if $f_A$,$f_B$ and $T$ are the frictional forces on block $A$,block $B$ and the tension in the string respectively,then their values are:

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