Question

Two rods of negligible weight are pinned at B and attached to blocks A and C. Find the value of P for impending motion and determine which block will move first, given: MA = 24 kg, MC = 60 kg, θ = 70 °, Φ = 50 °, α = 35 °, μ = 0.25

Answer 1

Answer:

$P = 253.476$ N

Block A will move first

Explanation:

Let's call:

$F_{AB}$ - the Force applied from A to B

$F_{CB}$ - the Force applied from C to B

$P$ - the Force applied by P

Since the directions of all Forces above are towards B:

Angles between $F_{AB}$ and $F_{CB}$  is $\Phi + \alpha = 50+35 = 85$°

Angles between $F_{CB}$ and $P$  is $\theta - \Phi = 70-50 = 20$°

Angles between $F_{AB}$ and $P$ is 180 - 85 - 20 = 75°

From Sine's Law:

$F_{CB} = Psin(75)/sin(85) = 0.9696P\\F_{AB} = Psin(20)/sin(85) = 0.3433P$

For Block A, from Newton's Law:

$\sum F_y = 0$ ⇒ $N_A - M_Ag -F_{AB}sin\alpha=0$

$N_A = 24*9.8 + 0.3433Psin(35) = 235.2 + 0.1969 P$

$\sum F_x = 0$ ⇒ $F_A - F_{AB}cos\alpha=0$ where $F_A = \mu N_A$

$F_A = 0.3433Pcos(35) = 0.2812 P$

Since $N_A = F_A/\mu$

Then $235.2 + 0.1969 P = 0.2812 P/0.25$ ⇒ $P = 253.476$ N

For Block B, from Newton's Law:

$\sum F_y = 0$ ⇒ $N_B - M_Bg -F_{CB}cos(90-\Phi)=0$

$N_B = 60*9.8 + 0.9696Pcos(40) = 588 + 0.7428 P$

$\sum F_x = 0$ ⇒ $F_{AB}sin(90-\Phi)-F_B=0$ where $F_B = \mu N_B$

$F_B= 0.9696Psin(40) = 0.6232 P$

Since $N_B = F_B/\mu$

Then $588 + 0.7428 P = 0.6232 P/0.25$ ⇒ $P = 336$ N

Hence,

Since the value of P is smaller for Block A than Block B, Block A will move first.

And the required value of P is smaller one among the above, $P = 253.476$ N