Question 34: Laws of Motion

Question

In the given figure the blocks $A, B$ and $C$ weigh $4 \mathrm{~kg}, 6 \mathrm{~kg}$ and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5 . The force $\vec{F}$ required to slide the block $C$ with constant speed is $\_\_\_\_$ N . (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

If Block C moves to the Left with velocity v. Then string pulls Block B to the Right with velocity v.
Since Block A is just sitting on Block B and the motion of block C is uniform motion so there will be kinetic friction between A and B .
For block $\mathrm{A}, \mathrm{N}_{\mathrm{AB}}-\mathrm{m}_{\mathrm{A}} \mathrm{g}=0 \Rightarrow \mathrm{~N}_{\mathrm{AB}}=\mathrm{m}_{\mathrm{A}} g$
So, the normal force between blocks A and B is $\mathrm{m}_{\mathrm{A}} \mathrm{g}=4 	imes 10 \mathrm{~N}=40 \mathrm{~N}$
So, the friction between A and B is, $f_{A B}=\mu N_{A B}=0.5 	imes 40 N=20 N \Rightarrow f_{A B}=20 N$
For block B,
$$ -\mathrm{N}_{\mathrm{BA}}+\mathrm{N}_{\mathrm{BC}}-\mathrm{m}_{\mathrm{B}} \mathrm{~g}=0 $$
$$\Rightarrow-\mathrm{m}_{\mathrm{A}} \mathrm{~g}+\mathrm{N}_{\mathrm{BC}}-\mathrm{m}_{\mathrm{B}} \mathrm{~g}=0 $$
$$\Rightarrow $$ $$ \mathrm{N}_{\mathrm{BC}}=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}ight) \mathrm{g} $$
So, the normal force between blocks B and C is $\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}ight) \mathrm{g}=(4+6) 	imes 10 \mathrm{~N}=100 \mathrm{~N}$
Block C moves left and B moves right. So, on block B friction acts to the left (opposing its rightward motion). And on block C friction acts to the right (opposing its leftward motion relative to B ).
$$ \mathrm{f}_{\mathrm{BC}}=0.5 	imes 100=50 \mathrm{~N} \Rightarrow \mathrm{f}_{B C}=50 \mathrm{~N} $$
Ground supports all the three blocks, so the normal force on the ground is -
$$ \mathrm{N}_{\mathrm{C}}=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{C}}ight) \mathrm{g} $$
$$\Rightarrow $$$$ N_C=(4+6+8) 	imes 10=18 	imes 10=180 N $$
Friction at ground ( $\mathrm{f}_{\mathrm{g}}$ ) acts on Block C to the Right (opposing motion).
$$ \mathrm{f}_{\mathrm{g}}=0.5 	imes 180=90 \mathrm{~N} \Rightarrow \mathrm{f}_{\mathrm{g}}=90 \mathrm{~N} $$
As block B is moving at constant speed, so the net force is zero.
$\mathrm{T}=\mathrm{f}_{A B}+\mathrm{f}_{\mathrm{BC}}$
$$\Rightarrow $$ $$ \mathrm{T}=20 \mathrm{~N}+50 \mathrm{~N}=70 \mathrm{~N} $$
Also block C is moving at constant speed, so the net force is zero.
$$ \mathrm{F}=\mathrm{T}+\mathrm{f}_{\mathrm{BC}}+\mathrm{f}_{\mathrm{g}} $$
$$\Rightarrow $$ $\mathrm{F}=70 \mathrm{~N}+50 \mathrm{~N}+90 \mathrm{~N}$
$$\Rightarrow $$ $$ \mathrm{F}=210 \mathrm{~N} $$
So, the force required is 210 N.

Laws of Motion

In the given figure the blocks A, B and C weigh 4 \mathrm{~kg}, 6 \mathrm{~kg} and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5 . The force \vec{F} required to slide the block C with constant speed is \_\_\_\_ N . (Use g=10 \mathrm{~m} / \mathrm{s}^2 )

Integer / Numeric Answer Type