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

`P = 253.476` N

Block A will move first

Step-by-step 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