Question

Block A has a mass of 14 kg, block B has a mass of 12 kg, and angle alpha equals 19°. What is the tension in the wire that connects block A to the wall?

Answer 1

83.0 N

Step-by-step explanation:

To solve this problem, We start by analyzing the forces acting on block B.

There are two forces acting on B along the direction of the plane:

- The tension in the wire,
`T_B`, acting up along the plane

- The component of the weight of the block,
`m_Bg sin \theta`, acting down along the plane, where mB is the mass of the block, g is the acceleration of gravity,
`\theta` is the angle of the incline

Therefore the equation of the forces on B along the direction of the plane is

`T_B-m_Bg sin \theta = ma`

where a is the acceleration. However, the block is in equilibrium, so a = 0 and the equation becomes

`T_B-m_Bg sin \theta = 0`

Substituting the data that we know:

mB = 12 kg (mass of block B)

`g=9.8 m/s^2`

`\theta = 19^(\circ)`

We can find the tension in the wire:

`T_B=m_Bg sin \theta = (12)(9.8)(sin 19^(\circ))=38.3 N`

Now we can write the equation of the forces acting on block A:

`T_A - m_A g sin \theta - T_B = 0`

where

`m_A = 14 kg` is the mass of block A

`T_A` is the tension in the wire connecting A with the wall

Solving for
`T_A`,

`T_A = m_A g sin \theta + T_B = (14)(9.8)(sin 19)+38.3=83.0 N`