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

Answer:

83.0 N

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 $T_A$,

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