High School

A 13 kg log is being dragged along flat ground by a 114 N force at an angle of 45 degrees. If the log accelerates at a rate of 0.72 m/s², which is the normal force acting on the log?

A. 124.0 N
B. 114.0 N
C. 96.0 N
D. 80.0 N

Answer :

Final answer:

To find the normal force acting on a log being dragged on flat ground, we consider the force's vertical component and subtract it from the gravitational force due to the log's weight. However, the calculated normal force does not align with the provided answer choices, suggesting a need to re-evaluate the problem or the answer options.

Explanation:

The question concerns determining the normal force acting on a log being dragged along flat ground with a known force at an angle. The log has a mass of 13 kg and is being pulled by a force of 114 N at a 45-degree angle, resulting in an acceleration of 0.72 m/s².

To find the normal force, we first break down the 114 N force into its components. The vertical component of this force can be found using trigonometry, specifically F_∇y = F × sin(θ), where F is the applied force and θ is the angle with the horizontal.

The vertical component of the force acting upwards is F_∇y = 114 N × sin(45°) = 80.6 N. The weight (W) of the log acts downwards and is given by W = m × g, where m is the mass and g is the acceleration due to gravity (9.81 m/s²). So, W = 13 kg × 9.81 m/s² = 127.53 N. The normal force (N) must counteract the weight minus the upward component of the applied force: N = W - F_∇y = 127.53 N - 80.6 N = 46.93 N.

However, since the question asks for the net normal force and there are no additional vertical forces given (such as additional downward forces), we should consider just the weight of the log and the force of gravity. Thus, the normal force is the gravitational force minus any vertical component of the pulling force.

Looking at the answer choices provided, none exactly match the calculated normal force of 46.93 N, which suggests a potential error in the calculation or the answer choices. Given the information provided, and without additional vertical forces, the complete free-body diagram needs to be considered with weight, normal force, friction, the vertical force due to the pulling force, and the resulting acceleration in the horizontal direction. Thus, without further information, we can state that the apparent normal force would be approximately 127.53 N, adjusting for the pulling angle.

However, since we need to find the normal force that acts on the log while accounting for the angled force, we add the vertical component back to the gravitational force to correct our calculation. This addition gives us: N = W + F_∇y = 127.53 N + 80.6 N = 208.13 N. We consider the angle's effect to be reducing the weight effect on the normal force. Therefore, the actual normal force, after subtracting the vertical component due to pulling, is approximately 127.53 N. Since this matches none of the answer choices, a re-evaluation of the problem or given answers may be necessary.

In summary, considering the physics of the problem, the normal force would be approximately 127.53 N before considering the pulling force, and after considering the pull and angle, it would still be approximately 127.53 N (if using the subtraction method, then 208.13 N), which unfortunately doesn't match any of the given choices. There may be additional information or instructions needed to determine the correct answer from the provided options.

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