4. Two girls hold a pleture against the wall to measure its height before their mother hammers a picture hook into the wall. One girl exerts a force of 35 t and the other girl exerts a force of 55 N . The angle between their arms is 90 a) Draw a free-body diagram of the forces acting on the picture. b) Draw a triangle of forces acting on the picture. c) Calculate the weight of the plcture. ( 65 N ) 5. The diagram shows the result from an experiment that determined the resultant of three non-linear forces acting on a point. The system is in equilibrium.

Let's tackle **Problem 4** step by step, focusing on part **c**, where we need to calculate the weight of the picture. However, I'll also briefly outline parts **a** and **b** for completeness. ### 4. Two girls hold a picture against the wall to measure its height before their mother hammers a picture hook into the wall. One girl exerts a force of 35 N and the other girl exerts a force of 55 N. The angle between their arms is 90 degrees. #### a) Free-Body Diagram of the Forces Acting on the Picture A free-body diagram (FBD) illustrates all the forces acting on an object. For the picture held against the wall: 1. **Downward Force (Weight, \( W \))**: This is the gravitational force acting on the picture. 2. **Upward Force by Girl 1 (\( F_1 = 35 N \))**: Exerted at one arm. 3. **Upward Force by Girl 2 (\( F_2 = 55 N \))**: Exerted at an arm perpendicular (90 degrees) to the first.  *Note: Since I cannot display images, imagine the picture with three vectors: \( W \) downward, \( F_1 \) upwards to the right, and \( F_2 \) upwards perpendicular to \( F_1 \).* #### b) Triangle of Forces Acting on the Picture In the triangle (or vector) diagram: 1. Draw \( F_1 \) horizontally. 2. From the tip of \( F_1 \), draw \( F_2 \) vertically upward. 3. The resultant vector (\( R \)), representing the weight \( W \), closes the triangle from the tip of \( F_2 \) back to the tail of \( F_1 \). This forms a right-angled triangle with \( F_1 \) and \( F_2 \) as the perpendicular sides and \( W \) as the hypotenuse.  *Again, imagine a right-angled triangle with sides 35 N and 55 N, and the hypotenuse representing \( W \).* #### c) Calculating the Weight of the Picture Since the system is in equilibrium (the picture is held stationary), the resultant of the two upward forces must balance the downward weight. To find the weight \( W \), we calculate the resultant of \( F_1 \) and \( F_2 \) using the Pythagorean theorem (because the forces are perpendicular to each other): \[ W = \sqrt{F_1^2 + F_2^2} \] Plugging in the given values: \[ W = \sqrt{(35 N)^2 + (55 N)^2} = \sqrt{1225 + 3025} = \sqrt{4250} \approx 65 N \] **Therefore, the weight of the picture is approximately 65 N.**