Question

if the box accelerates down the incline with magnitude a, what is the magnitude of the applied force f on the box?
choose 1 answer:
a ( m(a + gcos\theta) )
b ( m(a + gsin\theta) )
c ( m(a - gsin\theta) )
14 of 14

Explanation:

Step1: Analyze Forces on Incline

The box on an incline has gravitational force component \( mg\sin\theta \) down the incline (where \( \theta \) is the incline angle), applied force \( F \), and acceleration \( a \) down the incline. Using Newton's second law \( F_{\text{net}} = ma \), the net force down the incline is \( F + mg\sin\theta \) (assuming \( F \) is down the incline? Wait, no—wait, if acceleration is down, let's clarify: Let's assume the applied force \( F \) and the component \( mg\sin\theta \) are along the incline. Wait, maybe I got the direction wrong. Wait, actually, let's define the coordinate system along the incline (down as positive). The gravitational component down the incline is \( mg\sin\theta \), and let's say the applied force \( F \) is also down (or maybe up? Wait, the problem says "applied force \( F \)"—wait, maybe the force is applied, and we need to find \( F \) such that acceleration is \( a \) down. Wait, no—let's do Newton's second law properly.

Wait, let's correct: The net force along the incline is \( F_{\text{net}} = ma \). The forces along the incline: gravitational component \( mg\sin\theta \) (down), and applied force \( F \) (let's assume direction—wait, maybe the applied force is up? No, the acceleration is down, so net force is down. Wait, maybe the applied force is down, but let's re-express. Wait, no—let's suppose the applied force is \( F \), and the gravitational component is \( mg\sin\theta \) down. Then net force is \( F + mg\sin\theta = ma \)? No, that would be if both are down. But that would give \( F = ma - mg\sin\theta \), which is \( m(a - g\sin\theta) \)? Wait, no—wait, maybe I mixed up the direction. Wait, maybe the applied force is up, and the gravitational component is down. Then net force is \( mg\sin\theta - F = ma \) (if acceleration is down, then net force is down, so \( mg\sin\theta > F \), so \( F = mg\sin\theta - ma = m(g\sin\theta - a) \), but that's not an option. Wait, the options are \( m(a + g\cos\theta) \), \( m(a + g\sin\theta) \), \( m(a - g\sin\theta) \). Wait, maybe the angle is with respect to the vertical? No, incline angle is with respect to horizontal. Wait, no—wait, maybe the force is applied perpendicular? No, the problem says "applied force \( F \)" on the box, likely along the incline. Wait, maybe I made a mistake in the gravitational component. Wait, no—gravitational force component along incline is \( mg\sin\theta \), perpendicular is \( mg\cos\theta \). So along incline: forces are \( F \) (applied) and \( mg\sin\theta \) (gravitational), and net force is \( ma \). Wait, the options have \( g\sin\theta \), so let's re-arrange.

Wait, let's do Newton's second law: \( F_{\text{net}} = ma \). Let's assume the applied force \( F \) is in the direction opposite to the acceleration? No, the acceleration is down, so net force is down. Wait, maybe the applied force is up, and the gravitational component is down. Then net force is \( mg\sin\theta - F = ma \) (since acceleration is down, net force is down, so \( mg\sin\theta > F \)). Then solving for \( F \): \( F = mg\sin\theta - ma = m(g\sin\theta - a) \), which is \( m(-a + g\sin\theta) \), not matching. Wait, the options have \( m(a - g\sin\theta) \) as option C. Wait, maybe the applied force is down, and the gravitational component is up? No, that's not possible. Wait, maybe I got the direction of the applied force wrong. Let's re-express:

Suppose the applied force \( F \) is along the incline (let's say up), and the gravitational component is \( mg\sin\theta \) down. Then the net force is \( mg\sin\theta - F = ma \) (since acceleration is down, net force is down). Then solving for \( F \): \( F = mg\sin\theta - ma = m(g\sin\theta - a) = m(-a + g\sin\theta) \), which is not matching. Wait, the options have \( m(a - g\sin\theta) \) (option C), \( m(a + g\sin\theta) \) (option B), \( m(a + g\cos\theta) \) (option A). Wait, maybe the applied force is down, and the gravitational component is also down, but the acceleration is \( a \), so net force is \( F + mg\sin\theta = ma \)? No, that would give \( F = ma - mg\sin\theta = m(a - g\sin\theta) \), which is option C. Ah! That makes sense. So if the applied force \( F \) is down the incline, and the gravitational component \( mg\sin\theta \) is also down, then net force is \( F + mg\sin\theta = ma \)? Wait, no—wait, no, that would be if both are down, but then \( F = ma - mg\sin\theta = m(a - g\sin\theta) \), which is option C. So that's the correct approach.

So step 1: Identify forces along incline: \( F \) (applied, down), \( mg\sin\theta \) (gravitational, down), net force \( F + mg\sin\theta = ma \)? No, wait, no—net force is mass times acceleration, so \( F_{\text{net}} = ma \). If \( F \) and \( mg\sin\theta \) are both down, then \( F + mg\sin\theta = ma \) would imply \( F = ma - mg\sin\theta = m(a - g\sin\theta) \), which is option C.

Step2: Verify the Equation

Using Newton's second law \( F_{\text{net}} = ma \). The forces along the incline: gravitational component \( mg\sin\theta \) (down), applied force \( F \) (down). Then net force is \( F + mg\sin\theta = ma \)? No, that can't be, because if both are down, the net force would be their sum, but acceleration is \( a \), so \( F + mg\sin\theta = ma \) would mean \( F = ma - mg\sin\theta \), which is \( m(a - g\sin\theta) \), which is option C. So that's the correct derivation.

Answer:

C. \( m(a - g\sin\theta) \)