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Question

A $1130$ -kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E $5.8$ ). The cable makes an angle of $31.0°$ above the surface of the ramp, and the ramp itself rises at $25.0°$ above the horizontal. Draw a free-body diagram for the car.

An orange car on a frictionless ramp, held by a cable at a 31° angle, ramp inclined at 25°.

Accepted Answer

Hey everybody. So today we're dealing with the problem about force body diagrams. Now I've gone ahead and highlighted the important information in our question stem, but basically we're being told that a 1000 kg block is being held on a frictionless inclined wedge at an angle of 37 degrees above the horizontal. now the cable holding this block at this incline makes a 30° angle with the surface of the incline itself with this. We're being asked to sketch a free body diagram of the box. Now we have everything drawn here, but I'd like to make this a little simpler by drawing out the force body diagram with the horizontal being more uh or rather easier to see. So let's just say that that is our horizontal and we have a point right here, let us say this point is our block. No. What are the other things that we know? So let's write this, This is the horizontal. We have an incline. Oops, that'll do. For now we have an incline Of the 37° above the horizontal. This is 37°. We also have a tension force. And just to be clear this uh line that I've just drawn. This tension is the same as this uh cable that's drawn here is the source of tension and it is keeping the Box up and it forms an angle of 30° with the inclined plane, not with the horizontal, but with the incline itself. Now there's a couple other things, we need to remember. Normal force always acts perpendicular to a surface, which means the normal force here will be acting perpendicular to the surface on which the object is on which is the incline. So the normal force will be this way, this will be the normal force. The other thing that we know is that weight of a body will always act downwards. So let's draw that out. If it goes, if it acts downwards then that means that this would be MG. Now I'm just going to make our normal force extend a little further and you'll see why in just a second if we use our um rules of intersecting lines, we realize that this angle right here, He's also 37 degrees, which means we can actually use trig metric values to figure out what are the actual forces acting here because the force due to gravity or weight is acting purely downwards. But we're on an incline which means they're both horizontal and vertical components to this factor. And this makes sense. Right. We can see that if we have to split this into its component vectors, right? If this is the resultant force then there has to be a force this way and they force this way down the incline and almost sort of through the incline. Because when you add those up and remember we can add up vectors by just putting the points together that will give us the resultant MG. Or wait. And if you look closely this is a right triangle. So if we use our trick, no metric values or trick geometric rules, I should say. We can see that the downward force. Well it lies opposite to the, excuse me, it lies adjacent, It is lying adjacent to the Angle Theta or Angle 37. Which means that recalling our rules we have so car to her. If it is adjacent we use Kassian. So what this means the value in the X. Direction or this value right here. And let's draw that. Uh let's let's see screen for that. This value right here will be MG Kassian 37 because that is the angle in question 37 degrees. Similarly, if we use trigger magic rules again, this this leg of the triangle which is uh equivalent to this leg, it is it would be the same. This leg lies opposite to angle theta, which means if we use opposite then we need to use sign to calculate a value which means drawing it up here. This would equal The force of weight multiplied by a sign 37°. We can use the same logic now for the uh forces due to tension because look, tension is also acting at an angle right? Which means we have both a horizontal and vertical component. And this vertical component is essentially the same as the normal force. It is the normal force. In fact because if we add up the resultant vectors, if this is normal force and this is the other leg moving it up and adding them together, we will get the tension. So now using this angle, this is our angle and applying the same rules. Excuse me. The horizontal here. If we draw this out as a triangle and let me use let me go back here. If we draw this as a triangle, this leg lies adjacent to the angle data, which means like we did earlier, this will just be the force due to tension multiplied by cosign degrees because that is the angle in question. And similarly the normal force is lying opposite the angle theta, which means we use sign to calculate it, which means The normal force is equal to T sign 30°. So that will be our full force body diagram. I hope this helps. And I look forward to seeing you all in the next one.

A $1130$ -kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E $5.8$ ). The cable makes an angle of $31.0°$ above the surface of the ramp, and the ramp itself rises at $25.0°$ above the horizontal. Draw a free-body diagram for the car.

Key Concepts

Free-Body Diagram

Forces on an Incline

Tension in a Cable

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A 113011301130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.85.85.8). The cable makes an angle of 31.0°31.0°31.0° above the surface of the ramp, and the ramp itself rises at 25.0°25.0°25.0° above the horizontal. Draw a free-body diagram for the car.

Key Concepts

Free-Body Diagram

Forces on an Incline

Tension in a Cable

Watch next

A $1130$ -kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E $5.8$ ). The cable makes an angle of $31.0°$ above the surface of the ramp, and the ramp itself rises at $25.0°$ above the horizontal. Draw a free-body diagram for the car.