To find the tension in the second string and the weight of the picture, we can analyze the forces acting on the system. Let’s denote the tension in string 1 as T1 and the tension in string 2 as T2. We know that T1 = 20 N.
Since the picture is in equilibrium, the sum of the vertical forces must equal zero. The weight of the picture (W) acts downwards and is balanced by the tensions in the two strings (T1 and T2) acting upwards. Thus, we can express this relationship as:
W = T1 + T2
Next, to find T2, we need more information regarding the angles at which the strings are positioned or any additional conditions specified. However, if we assume that both strings are oriented such that T1 and T2 contribute equally to supporting the weight, we can simplify the situation.
If T1 holds 20 N and there is an assumption of symmetry (meaning T1 = T2), we would have:
T2 = T1 = 20 N
Thus, under these assumptions:
W = 20 N + 20 N = 40 N
In conclusion, assuming equal distribution, the tension in the second string (T2) is 20 N, and the weight of the picture (W) is 40 N.