College

A 10 kg mass is suspended by two strings, as shown below.

M

20°

F

F

20°

10 kg mass

The force applied to the mass by each of the strings, F is closest to:

(A) 98 N

(B) 143 N

(C) 52 N

(D) 72 N

A 10 kg mass is suspended by two strings as shown below M 20 F F 20 10 kg mass The force applied to the

Answer :

Option B. The force applied by each of the strings is approximately 143.3 N.

The mass is 10 kg, and gravity acts downward, applying a force equal to the weight of the mass:

W = mg = 10bkg * 9.8 m/s² = 98 N

This weight is balanced by the vertical components of the tension forces from both strings.

Since the mass is in equilibrium, the sum of the forces in both the vertical and horizontal directions must be zero:

Vertical direction (y-axis): The vertical components of the tension from each string must add up to balance the weight of the mass.

2F sin(20⁰) = 98 N

where F is the tension in each string, and sin(20⁰) is the vertical component of the tension force.

The horizontal components of the tension forces from each string cancel each other out because the mass is not moving sideways:

F cos(20⁰) - F cos(20⁰) = 0

Using the vertical equilibrium equation:

2F sin(20⁰) = 98

Solve for F

F = [tex]\frac{98}{2 \sin(20^\circ)}[/tex]

F =[tex] \frac{98}{2 \times 0.342}[/tex]

= 98/0.684 = 143.3 N