FSQ TPR Physics Incline planes

[Futbol99]

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Question:
A 40-kg crate is being pulled along a frictionless surface by a force of magnitude 140N that makes an angle of 30 degrees with the horizontal. What is the acceleration of the crate?

A. 1.75
B. 2
C. 2.5
D. 3

I know that this question was posted before once on this forum a long time ago, but I didn't get their explanation. I don't understand the books explanation either.
I thought it was a regular incline type of question, where its: 140Sin30/40 which gets 1.75. But the book explains using 140cos30/40, and says "The crate can only move in the horizontal direction, we only consider the horizontal component of the applied force".

I understand how the answer would differ if the crate can only move in the horizontal direction, but HOW CAN I ASSUME THAT about the crate?
Almost every incline question I do and get right uses the same words in the problem. How do I assume it only moves horizontally based on saying "makes an angle of 30 deg with the horizontal"....

Please help, very confused 🙂

 

[Cmdr_Shepard]

Well, the question stem says "pulled along a frictionless surface", so from that you can assume it's moving horizontally.

Also, just from calculating Fsin30, you can see that the upward force is not enough to overcome gravity and accelerate the crate vertically at all.

Horizontally, you've got Fcos30 = ma. Plug in 40 kg and 140 N and solve for a.

 

[Futbol99]

Well, the question stem says "pulled along a frictionless surface", so from that you can assume it's moving horizontally.

Also, just from calculating Fsin30, you can see that the upward force is not enough to overcome gravity and accelerate the crate vertically at all.

Horizontally, you've got Fcos30 = ma. Plug in 40 kg and 140 N and solve for a.

hmm..

I just looked at another question, that uses similar words that I got right, but I guess a small distinction can be made, so let me know if this is what you meant.

So this question says "pulled along a frictionless surface" and so that assumes we move it horizontally, thus use the Cos component.
The other question uses Sin component says "Block slides down a frictionless surface".
So due to the word usage of "down" that means use the vertical component?

Also could you elaborate on that second part about overcoming gravity? I'm trying to understand this chapter better.

Thanks

 

[Cmdr_Shepard]

Well it sounds like you're also talking about inclined planes. So I'd recommend drawing free-body diagrams for each question and then applying F=ma to the diagram.

Your FBD for the first question would have the angled force, gravity, and the normal force all acting on the crate. Gravity, the vertical component of the force, and the normal force would balance to give zero accel in the vertical direction. The Fcos30 would be the only force acting in the horizontal direction and according to F = ma, would accelerate the crate.

For a block on an inclined plane, you have to draw a FBD to take into account parallel and perpendicular forces on the crate-plane system. You can use terminology like "down the plane", "up the plane", "normal to the plane", etc if that makes you understand it better. What you would have on a friction-less plane is mg pointing straight down, and the normal force acting perp. to the surface of the incline. The Normal does not have to be resolved into its components to solve the question, but the mg does. You would break it up into parallel and perp. forces. The parallel force is Fsin(theta), and the perp. one is Fcos(theta), which is also equal to the Normal force since there is no acceleration in the perp. direction. Thus, the Fsin(theta) is the only force accelerating the block down the incline. Then use F=ma to solve.

The gravity bit is just basically saying that the weight of the block is too much for Fsin30 to overcome, which leads to zero accel. in the vertical direction.

 

[Futbol99]

Well it sounds like you're also talking about inclined planes. So I'd recommend drawing free-body diagrams for each question and then applying F=ma to the diagram.

Your FBD for the first question would have the angled force, gravity, and the normal force all acting on the crate. Gravity, the vertical component of the force, and the normal force would balance to give zero accel in the vertical direction. The Fcos30 would be the only force acting in the horizontal direction and according to F = ma, would accelerate the crate.

For a block on an inclined plane, you have to draw a FBD to take into account parallel and perpendicular forces on the crate-plane system. You can use terminology like "down the plane", "up the plane", "normal to the plane", etc if that makes you understand it better. What you would have on a friction-less plane is mg pointing straight down, and the normal force acting perp. to the surface of the incline. The Normal does not have to be resolved into its components to solve the question, but the mg does. You would break it up into parallel and perp. forces. The parallel force is Fsin(theta), and the perp. one is Fcos(theta), which is also equal to the Normal force since there is no acceleration in the perp. direction. Thus, the Fsin(theta) is the only force accelerating the block down the incline. Then use F=ma to solve.

The gravity bit is just basically saying that the weight of the block is too much for Fsin30 to overcome, which leads to zero accel. in the vertical direction.

Awesome man,

Thank you so much. I guess the wording in this question just got me really tripped up! But yeah all my questions were about the inclined planes, which you solidified for me!

Thanks again 🙂

 

[Cmdr_Shepard]

No prob!

I guess being an MCAT instructor for 4 years (and counting) has taught me how to explain stuff!