Question

# A block A of mass 100kg rests on another block B of mass 200kg and is tied to a wall. The coefficient of friction between A And B is 0.2 and that between B and the ground is 0.3. What will be the minimum force F required to move the block B? (G=10m/S 2)

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Solution

## Step1: Given dataMass of block A =${m}_{A}=100kg$Mass of block B = ${m}_{B}=200kg$Coefficient of friction between A and B = ${\mu }_{1}=0.2$Coefficient of friction between B and ground = ${\mu }_{2}=0.3$Step2: Coefficient of frictionThe coefficient of friction between two surfaces represents the amount of interaction between the two surfaces in terms of friction. The coefficient of friction value can be used to determine surface roughness. When any object under observation is subjected to the normal force operating on it, friction occurs. Step3: Drawing a free body diagramIn the given question there are so many forces like,The friction force between block A and block B = ${f}_{1}$The friction force between block B and block ground = ${f}_{2}$Tension in the rope = $T$Step4: Formula usedThe frictional force is calculated as, $f=\mu mg\phantom{\rule{0ex}{0ex}}\left[where\mu =coefficientoffriction,\phantom{\rule{0ex}{0ex}}m=massofobject,\phantom{\rule{0ex}{0ex}}g=accelerationduetogravity\right]$Step5: Calculating frictional force The frictional force acting between blocks A and B ${f}_{1}={\mu }_{1}{m}_{A}g=0.2×100×10=200N$The frictional force acting between block B and ground ${f}_{2}={\mu }_{2}{m}_{B}g=0.3×\left(100+200\right)×10=900N$Step6: Calculating minimum frictional forceNow, we can see from the diagram that the minimum force required to move block B will be sufficient to overcome the friction forces (friction forces between both blocks and friction forces between the ground and block A). As a result, the minimum value must be the total of both friction forces.Hence the force required to move the block $F={f}_{1}+{f}_{2}\phantom{\rule{0ex}{0ex}}F=200+900\phantom{\rule{0ex}{0ex}}F=1100N$Hence the required force will be 1100N.

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