Question

A block having a mass of m = 19.5 kg is suspended via two cables as shown in the figure. The angles shown in the figure are as follows: a = 15° and B = 35°. We will label the tension in Cable 1 as T1, and the tension in Cable 2 as T2. Solve for T1 and T2

Answer 1

The tension in cable 2 supporting the hanging mass is determined as 109.6 N.

How to calculate the tension in cable 2?

The tension in cable 2 is calculated by applying the following formula as shown below;

T₂ = W sinβ

where;

• W is the weight of the hanging mass
• β is the angle of inclination of the cable 2

T₂ = mg sinβ

where;

• m is the mass of the block
• g is acceleration due to gravity

The tension in cable 2 is calculated as follows;

T₂ = mg sinβ

T₂ = (19.5 kg x 9.8 m/s²) x sin (35)

T₂ = 109.6 N

Answer 2

We will have the next diagram

Then we can sum the forces in x and sum the forces in y

Forces in x

`\sum ^{}_{}F_x=-T_1\sin (\alpha)+T_2\cos (\beta)=0`
`\sum ^{}_{}F_x=-T_1\sin (15)+T_2\cos (35)=0`

Forces in y

`\sum ^{}_{}F_y=T_1\cos (\alpha)+T_2\sin (\beta)=mg`
`\sum ^{}_{}F_y=T_1\cos (15)+T_2\sin (35)=19.5(9.8)`

We simplify the equations found and we found the next system of equation

`\begin{gathered} -T_1\sin (15)+T_2\cos (35)=0 \\ T_1\cos (15)+T_2\sin (35)=191.1 \end{gathered}`

then we isolate the T2 of the first equation

`T_2=(T_1\sin(15))/(\cos(35))`

We substitute the equation above in the second equation

`T_1\cos (15)+((T_1\sin(15))/(\cos(35)))\sin (35)=191.1`

we simplify

`T_1(\cos (15)+(\sin (15)\sin (35))/(\cos (35)))=191.1`
`T_1(1.147)=191.1`

We isolate the T1

`T_1=\frac{191.1}{1.147^{}}=166.6N`

then we can substitute the value we found in T1 in the euation with T2 isolate

`T_2=\frac{(166.6)_{}\sin (15)}{\cos (35)}=52.54N`