Question

The system of two blocks, cable and fixed pulley is initially at rest Determine the horizontal force (P) necessary to cause motion then determine the corresponding tension (T) in the cable​

Answer 1

Given that,

Mass of small block = 5 kg

Mass of big block = 10 kg

Suppose, P is applied to 5 kg block.

For small block.

For vertically,

We need to calculate the normal force

Using balance equation

$\sum{F_{y}}=0$

$N_{1}-mg=0$

Put the value into the formula

$N_{1}-(5\times9.8)=0$

$N_{1}=49.0\ N$

For horizontally,

Again using balance equation

$\sum{F_{x}}=0$

$P-F_{1}-T=0$.....(I)

For big block,

For vertically,

We need to calculate the normal force

Using balance equation

$\sum{F_{y}}=0$

$N_{2}-N_{1}-mg=0$

Put the value into the formula

$N_{2}-49-10\times9.8=0$

$N_{2}=147\ N$

For horizontally,

Again using balance equation

$\sum{F_{x}}=0$

$F_{1}+F_{2}-T=0$.....(II)

For the block to cause motion

We need to calculate the force F₁

Using formula of force

$F_{1}=F_{max}$

$F_{1}=\mu N_{1}$

Put the value into the formula

$F_{1}=0.5\times49$

$F_{1}=24.5\ N$

We need to calculate the force F₂

Using formula of force

$F_{2}=F_{max}$

$F_{2}=\mu N_{2}$

Put the value into the formula

$F_{2}=0.6\times147$

$F_{2}=88.2\ N$

We need to calculate the value of tension in the cable

Using equation (II)

$T=F_{1}+F_{2}$

Put the value of F₁, F₂ in equation (II)

$T=24.5+88.2$

$T=112.7\ N$

We need to calculate the value of P

Using equation (I)

$P=F_{1}+T$

Put the value in the equation

$P=24.5+112.7$

$P=137.2\ N$

Hence, The force P corresponding tension in the cable is 137.2 N.