Question

two forces are applied to a car in an effort to move it, as shown in the picture. what is the resultant of these two forces? what direction is the force relative to mathematical 0° (assume the car is pointing towards mathematical 0°)? if the car remains stationary, what is the magnitude and direction of the friction produced by the wheels? resultant 799 n @ × equilibrant 799 n @ 171 check

Explanation:

To solve the problem, we analyze the forces acting on the car:

Step 1: Resultant Force

When two forces act on an object, the resultant force is the vector sum. If the car is stationary (or moving at constant velocity), the resultant force is balanced by friction. However, the problem first asks for the resultant of the two applied forces. From the given data, the resultant magnitude is \( 799 \, \text{N} \), and the direction (relative to \( 0^\circ \)) is determined by the force diagram (not fully visible, but the equilibrant and resultant are related).

Step 2: Friction (Equilibrant Force)

If the car remains stationary, the friction force (equilibrant) must balance the resultant force. The equilibrant has the same magnitude as the resultant but the opposite direction. From the given equilibrant data, the magnitude of friction (equilibrant) is \( 799 \, \text{N} \), and its direction is opposite to the resultant (e.g., if the resultant is at \( \theta \), friction is at \( \theta + 180^\circ \), or relative to \( 0^\circ \), it would be \( 171^\circ \) as per the diagram’s context).

Final Answers

• Resultant Force: Magnitude \( \boldsymbol{799 \, \text{N}} \), Direction (relative to \( 0^\circ \)): Depends on the force diagram (e.g., \( \boldsymbol{171^\circ} \) if the resultant is at \( 171^\circ \) or similar).
• Friction (Equilibrant): Magnitude \( \boldsymbol{799 \, \text{N}} \), Direction opposite to the resultant (e.g., \( \boldsymbol{171^\circ} \) if the resultant is at \( 171^\circ \), or adjusted based on the diagram).

(Note: The exact direction relies on the force diagram’s orientation, but the magnitude of friction equals the resultant force for equilibrium.)

Answer:

To solve the problem, we analyze the forces acting on the car:

Step 1: Resultant Force

When two forces act on an object, the resultant force is the vector sum. If the car is stationary (or moving at constant velocity), the resultant force is balanced by friction. However, the problem first asks for the resultant of the two applied forces. From the given data, the resultant magnitude is \( 799 \, \text{N} \), and the direction (relative to \( 0^\circ \)) is determined by the force diagram (not fully visible, but the equilibrant and resultant are related).

Step 2: Friction (Equilibrant Force)

If the car remains stationary, the friction force (equilibrant) must balance the resultant force. The equilibrant has the same magnitude as the resultant but the opposite direction. From the given equilibrant data, the magnitude of friction (equilibrant) is \( 799 \, \text{N} \), and its direction is opposite to the resultant (e.g., if the resultant is at \( \theta \), friction is at \( \theta + 180^\circ \), or relative to \( 0^\circ \), it would be \( 171^\circ \) as per the diagram’s context).

Final Answers

• Resultant Force: Magnitude \( \boldsymbol{799 \, \text{N}} \), Direction (relative to \( 0^\circ \)): Depends on the force diagram (e.g., \( \boldsymbol{171^\circ} \) if the resultant is at \( 171^\circ \) or similar).
• Friction (Equilibrant): Magnitude \( \boldsymbol{799 \, \text{N}} \), Direction opposite to the resultant (e.g., \( \boldsymbol{171^\circ} \) if the resultant is at \( 171^\circ \), or adjusted based on the diagram).

(Note: The exact direction relies on the force diagram’s orientation, but the magnitude of friction equals the resultant force for equilibrium.)