Forces and equilibrium questionWatch

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Thread starter 6 years ago
#1
Just wanted to make sure on my method for the first question, because the force F is going across is it 3.4sin50? Because its 25 degrees more. And the normal reaction is 3.4cos25?

2nd question I realise that a line can be drawn across and down to make an x and y axis and resolve, but the question asks to draw to scale I don't understand how long to draw.

Any help would be appreciated 1
6 years ago
#2
(Original post by raiden95)
Just wanted to make sure on my method for the first question, because the force F is going across is it 3.4sin50? Because its 25 degrees more. And the normal reaction is 3.4cos25?

2nd question I realise that a line can be drawn across and down to make an x and y axis and resolve, but the question asks to draw to scale I don't understand how long to draw.

Any help would be appreciated umm. When I look at the question, I'm a little confused as to why they ask for the normal reaction first. I would have thought you'd need the horizonal force first. Unless I'm getting mixed up with this stuff...

I thought to get horizonal force, F you do 3.4sin25 divided by cos25, to get 1.59? Then normal reaction is 3.4cos25 - 1.59sin25?
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Thread starter 6 years ago
#3
(Original post by Pride)
umm. When I look at the question, I'm a little confused as to why they ask for the normal reaction first. I would have thought you'd need the horizonal force first. Unless I'm getting mixed up with this stuff...

I thought to get horizonal force, F you do 3.4sin25 divided by cos25, to get 1.59? Then normal reaction is 3.4cos25 - 1.59sin25?
Why is horizontal, 3.4sin25 divided by cos25, and I thought normal is the same as the force downwards as its not moving. But i think I'm wrong because the force is horizontal and not horizontal to the plane

EDIT: I have not yet been taught the method you say, maybe there's more than one way of doing the question?
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6 years ago
#4
(Original post by raiden95)
Why is horizontal, 3.4sin25 divided by cos25, and I thought normal is the same as the force downwards as its not moving. But i think I'm wrong because the force is horizontal and not horizontal to the plane

EDIT: I have not yet been taught the method you say, maybe there's more than one way of doing the question?
I did that because it says the truck isn't moving, meaning the component of the horizontal force parallel to the slope is equal to the component of the truck's weight parallel to the slope.
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Thread starter 6 years ago
#5
(Original post by Pride)
I did that because it says the truck isn't moving, meaning the component of the horizontal force parallel to the slope is equal to the component of the truck's weight parallel to the slope.
Im really lost, why there has to be division. I want to understand the maths behind it, can you explain ?
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Thread starter 6 years ago
#6
Bump
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6 years ago
#7
(Original post by raiden95)
Im really lost, why there has to be division. I want to understand the maths behind it, can you explain ? Ok

Assuming the ground is smooth, so there's no friction, the only forces are the weight of the truck, the horizontal force, and the normal reaction.

The question says the truck isn't moving. This means the components parallel to the slope must be equal. So you know that the horizontal force component is equal to the force going down the hill due to the weight, 3.4sin25. So you have the component of F. To find the force you divide it by cos25. It should make sense if you look at the diagram.

Then using F, you know that N must be equal to the components of the other forces perpendicular to the slope, because again the truck isn't moving. So N = 3.4cos25 + Fsin25
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Thread starter 6 years ago
#8
(Original post by Pride) Ok

Assuming the ground is smooth, so there's no friction, the only forces are the weight of the truck, the horizontal force, and the normal reaction.

The question says the truck isn't moving. This means the components parallel to the slope must be equal. So you know that the horizontal force component is equal to the force going down the hill due to the weight, 3.4sin25. So you have the component of F. To find the force you divide it by cos25. It should make sense if you look at the diagram.

Then using H, you know that N must be equal to the components of the other forces perpendicular to the slope, because again the truck isn't moving. So N = 3.4cos25 + Fsin25
Can you check my working please
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6 years ago
#9
(Original post by raiden95)
Can you check my working please
I think that looks good.
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