Two masses A and B of $$5$$ kg and $$6$$ kg respectively are connected by a string passing over a frictionless pulley fixed at the corner of table as shown in the figure. The coefficient of friction between A and the table is $$0.3$$. Find the minimum mass of C (in$$ $$kg) that must be placed on A to prevent it from moving (Take$$ g $$=$$ $$10$$ $$m/s2).

Question

Infinity Learn · Accepted Answer

$$T=60NT=$$μ$$N($$ To prevent A from moving $$)T=$$μ(50+10m)⇒μ$$(50+10m)=60$$⇒$$50+10m=600.3=200$$⇒$$m=15kg$$

Friction

Remember concepts with our Masterclasses.

$\begin{array}{l} T = 60 N \\ T = μN (To prevent A from moving) \\ T = μ (50 + 10 m) \\ \Rightarrow μ (50 + 10 m) = 60 \\ \Rightarrow 50 + 10 m = \frac{60}{0 .3} = 200 \\ \Rightarrow m = 15 kg \end{array}$

Ready to Test Your Skills?

free study material

Friction

Remember concepts with our Masterclasses.

T=60NT=μN( To prevent A from moving )T=μ(50+10m)⇒μ(50+10m)=60⇒50+10m=600.3=200⇒m=15kg

Ready to Test Your Skills?

free study material

$\begin{array}{l} T = 60 N \\ T = μN (To prevent A from moving) \\ T = μ (50 + 10 m) \\ \Rightarrow μ (50 + 10 m) = 60 \\ \Rightarrow 50 + 10 m = \frac{60}{0 .3} = 200 \\ \Rightarrow m = 15 kg \end{array}$