Given:
The mass of an object is 900 kg.
The angle made by string 1 with the horizontal right is 40°.
The angle made by string 2 with the horizontal left is 20°.
Standard data: The acceleration due to gravity is 9.8 m/s^2.
To find:
The tension in the string 1.
Explanation:
Let, the tension in string 1 be T1 and the tension in string 2 be T2. The tension in both the string is resolved in their horizontal and vertical components as shown in the diagram.
The weight of the object mg acts vertically downwards.
The horizontal components of the string are acting in opposite directions and thus balance each other. Thus, we get:
[tex]T_1cos40\degree=T_2cos20\degree[/tex]
Rearranging the above equation, we get:
[tex]T_2=\frac{T_1cos40\degree}{cos20\degree}\text{ ..... \lparen1\rparen}[/tex]
The vertically upwards sine components of the tension add up which balances the vertically downwards acting weight mg of the object. Thus, we get:
[tex]T_1sin40\degree+T_2sin20\degree=mg[/tex]
Substituting T2 from equation (1) in the above equation, we get:
[tex]\begin{gathered} T_1sin40\degree+\frac{T_1cos40\degree sin20\degree}{cos20\degree}=mg \\ \\ T_1(sin40\degree+\frac{cos40\degree sin20\degree}{cos20\degree})=mg \\ \\ T_1=\frac{mg}{(sin40\degree+\frac{cos40\degree sin20\degree}{cos20\degree})} \end{gathered}[/tex]
Substituting the values in the above equation, we get:
[tex]\begin{gathered} T_1=\frac{900\text{ kg}\times9.8\text{ m/s}^2}{(s\imaginaryI n40\operatorname{\degree}+\frac{cos40\operatorname{\degree}s\imaginaryI n20\operatorname{\degree}}{cos20\operatorname{\degree}})} \\ \\ T_1=\frac{8820\text{ kg m/s}^2}{0.6428+\frac{0.7660\times0.3420}{0.9397}} \\ \\ T_1=\frac{8820(\text{kgm\/s})^2}{0.6428+0.2788} \\ \\ T_1=9570.312\text{ N} \end{gathered}[/tex]
Option (D) 9570.261 N is very close to 9570.312 N.
Final Answer:
The tension in string 1 is 9750.260 N. Thus, the correct option is D.
