(Physics) Picture suspended on wall by two strings

Question

A picture on a wall is hung up by two strings. String 1 exerted a force T1 with a magnitude of 2.4 N at an angle of 73 degrees above the horizontal. String 2 exerts a force T2 at angle of 47 degrees above the horizontal. The magnitude of the gravitational acceleration is 9.8 m/s^2.

A) What is the magnitude of a force T2 in string 2?

B) What is the mass of the picture?

Answer 1

Let

T1 = 2.4 N be the force of the first string,

T2 (to be calculated) be the force of the second string,

a1 = 73° be the angle of the first string,

a2 = 47° be the angle of the second string,

g = 9.8 m/s² be gravitational acceleration,

m (to be calculated) be the mass of the picture,

F = mg be the weight of the picture.

I assume that we are to assume that the picture is stable.

That implies that the vector sum

T1 + T2 + F = 0

That implies that the two angles a1 and a2 are in opposite direction.

The sum of the three forces continues to be 0 even when projected on any axis.

It is convenient to project it on the horizontal and on the vertical axis.

The projection on the horizontal axis is calculated using the cosines of the angles involved,

and the projection on the vertical axis is governed by sines.

We can calculate T2 from horizontal projection because the horizontal projection of the

unknown force F is 0.

T1cos(a1) = T2cos(a2)

T2 = T1cos(a1)/cos(a2)

Substituting actual numbers

T2 = 2.4 × cos(73°) / cos(47°) = 1.03 N

We can calculate F and m from vertical projection

T1sin(a1) + T2sin(a2) = mg

m = (T1sin(a1) + T2sin(a2))/g

Substituting actual numbers

m = (2.4×sin(73°) + 1.03×sin(47°))/9.8 = 0.31 kg