Question

the free body diagram shows a box being pulled to the left up a 25 - degree incline. the magnitude of the normal force is n.

Answer:

1. Explanation:

• Step 1: Analyze forces perpendicular to the incline
• Assume the incline - plane is the \(x - y\) plane with the \(y\) - axis perpendicular to the incline. We need to find the net force in the direction perpendicular to the incline. Let's consider the forces acting on the box. The forces given are \(75N\), \(45N\), \(20N\), and \(50N\). We need to resolve the forces in the direction perpendicular to the incline. However, since no information about the orientation of these forces with respect to the incline is given other than the general statement of the box being pulled up a 25 - degree incline, we assume that the \(50N\) force is the weight - like force acting vertically downwards. We resolve this vertical force into components along and perpendicular to the incline. The component of the vertical force perpendicular to the incline is \(F_{perp}=50\cos(25^{\circ})\). Also, the \(45N\) force seems to have a component in the perpendicular direction. If we assume the \(45N\) force is acting at an angle to the incline such that its perpendicular component is considered. Let's assume the forces are in a 2 - D plane.
• First, resolve the vertical force \(F = 50N\) perpendicular to the incline. Using the formula \(F_{y}=F\cos\theta\) (where \(\theta = 25^{\circ}\)), we have \(F_{1}=50\cos(25^{\circ})\approx50\times0.9063 = 45.315N\).
• Now, assume the \(45N\) force has a component perpendicular to the incline. If we assume it is acting at an angle \(\alpha\) to the incline, and for simplicity, if we assume it is acting at an angle such that its perpendicular component is \(F_{2}\). Let's assume it is acting at an angle such that its perpendicular component is \(45\sin(25^{\circ})\approx45\times0.4226 = 19.017N\) (assuming a proper orientation).
• The normal force \(N\) balances the net force in the perpendicular direction.
• The net force in the perpendicular direction is the sum of the perpendicular components of the forces acting on the box.
• \(N = 50\cos(25^{\circ})-45\sin(25^{\circ})\) (assuming the direction of the \(45N\) force reduces the normal - like effect).
• \(N=45.315 - 19.017=26.298N\approx26.3N\). But if we assume the \(45N\) force has a component that adds to the normal force (a more appropriate assumption if it is a pulling force with a perpendicular - to - incline component), we consider the forces in the perpendicular direction.
• The vertical force of \(50N\) has a perpendicular component \(F_{perp1}=50\cos(25^{\circ})\approx45.32N\). If the \(45N\) force has a perpendicular component that adds to the normal force, and assuming the other forces \(75N\) and \(20N\) have no contribution in the perpendicular direction (since no information about their orientation is given), the normal force \(N\) is the sum of the perpendicular components of the forces acting on the box.
• The normal force \(N = 50\cos(25^{\circ})+45\sin(25^{\circ})\)
• \(N\approx45.32 + 19.02=64.34N\)
• Let's assume the forces are acting in a way that the normal force is the sum of the perpendicular components of the non - parallel to the incline forces.
• The normal force \(N\) is the sum of the perpendicular components of the forces. The vertical force of \(50N\) gives a perpendicular component \(F_{v\perp}=50\cos(25^{\circ})\approx45.32N\). If we assume the \(45N\) force has a perpendicular component \(F_{45\perp}=45\sin(25^{\circ})\approx19.02N\)
• \(N=50\cos(25^{\circ})+45\sin(25^{\circ})\)
• \(N\approx45.32+19.02 = 64.34N\approx64.3N\)

1. Answer:

• \(64.3\) (rounded to one decimal place)