A particle P of weight 20 N is suspended by two strings AP and BP. The tension in string AP is (ai+8j) and the tension in string BP is (6i+bj)

The value of a is?

And the value of b is?

The magnitude of the tension in string AP is?

The angle that string BP makes with the vertical is?

Give your answers as exact whole numbers or as decimals to 3 s. f.

i. a = -6 ii. b = 12 iii. 100 iv. 26.6°

Step-by-step explanation:

Since the weight of the object is 20 N and directed downwards, its vector is -20j. Now, since the particle is in suspension, AP + BP = -20j

So, since AP = ai + 8j and BP = 6i + bj,

AP + BP = -20j

ai + 8j  + 6i + bj = -20j

ai + 6i + 8j + bj = -20j

(a + 6)i + (8 + b)j = -20j

(a + 6)i + (8 + b)j = 0i - 20j

equating the components on both sids,

a + 6 = 0 and 8 + b = 20

a = -6 and b = 20 - 8 = 12

i. The value of a is -6

ii. The value of b is 12

iii. The magnitude of the tension in string AP is?

since AP = ai + 8j  = -6i + 8j

magnitude of AP = √[-6² + 8²] = √[36 + 64] = √100 = 10

iv. The angle that string BP makes with the vertical is?

Since BP = 6i + bj = 6i + 12j, the angle it makes with the horizontal is Ф = tan⁻¹(12/6) = tan⁻¹(2) = 63.44°.

Thus, the angle it makes with the vertical is thus 90° - 63.44° = 26.56° (complementary angles) ≅ 26.6° to 3 s.f

replication and blocking

step-by-step explanation: