Three forces are acting on a body. Two of the forces are 10 N due East and 5 N at 60 degrees North-East. If the body is in equilibrium, what is the third force?

Morning mercyjoe8687 — let's get this done. Step 1: Define the forces and the equilibrium condition. Let F_1, F_2, and F_3 be the three forces. Given: F_1 = 10\,N due East. F_2 = 5\,N at 60^ North-East. Since the body is in equilibrium, the vector sum of the forces is zero: F_1 + F_2 + F_3 = 0 Therefore, the third force F_3 is the negative of the resultant of the first two forces: F_3 = -(F_1 + F_2) Step 2: Resolve the known forces into their x and y components. For F_1 (10\,N due East, angle 0^): F_1x = 10 (0^) = 10 × 1 = 10\,N F_1y = 10 (0^) = 10 × 0 = 0\,N For F_2 (5\,N at 60^ North-East, angle 60^): F_2x = 5 (60^) = 5 × 0.5 = 2.5\,N F_2y = 5 (60^) = 5 × sqrt(3)2 ≈ 5 × 0.8660 = 4.330\,N Step 3: Calculate the components of the resultant force of F_1 and F_2. Let R = F_1 + F_2. R_x = F_1x + F_2x = 10\,N + 2.5\,N = 12.5\,N R_y = F_1y + F_2y = 0\,N + 4.330\,N = 4.330\,N Step 4: Determine the components of the third force F_3. Since F_3 = -R: F_3x = -R_x = -12.5\,N F_3y = -R_y = -4.330\,N Step 5: Calculate the magnitude of the third force. The magnitude F_3 is given by: F_3 = sqrt(F_3x)^2 + F_3y^2 F_3 = sqrt((-12.5\,N))^2 + (-4.330\,N)^2 F_3 = sqrt(156.25\,N)^2 + 18.7489\,N^2 F_3 = sqrt(174.9989\,N)^2 F_3 ≈ 13.228\,N Step 6: Calculate the direction of the third force. The direction _3 is found using the tangent function. First, find the reference angle : = | F_3yF_3x | = | -4.330\,N-12.5\,N | = (4.330)/(12.5) ≈ 0.3464 = (0.3464) ≈ 19.1^ Since both F_3x and F_3y are negative, the third force lies in the third quadrant. The angle measured counter-clockwise from the positive x-axis is: _3 = 180^ + = 180^ + 19.1^ = 199.1^ Alternatively, this direction can be described as 19.1^ South of West. The magnitude of the third force is approximately 13.2 N and its direction is approximately 199.1^ from the positive x-axis (or 19.1^ South of West). Send me the next one 📸