How to find acceleration and tension given mass and an angle

In summary, a mass of 65.00 kg is suspended by a massless string from the ceiling of a van which is moving with constant acceleration a. The string makes an angle of 26.00 degrees with respect to the vertical. To find the acceleration of the van, you would need to consider the forces acting on the mass, including the weight of the object and the tension in the string. Using the equation \Sigma F = ma, you can set the sum of the forces in the x and y directions equal to the mass times acceleration. This will allow you to solve for the acceleration of the van. To find the tension in the string, you will need to use the formula for the magnitude of the tension, which is equal
  • #1
angiep410
39
0
A mass M = 65.00 kg is suspended by a massless string from the ceiling of a van which is moving with constant acceleration a. If the string makes an angle theta = 26.00 degrees with respect to the vertical:
a.what is the acceleration a of the van?
b.What is the tension in the string?

can someone please explain this to me
 
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  • #2
For part (a) think about the forces involved as sides of a triangle. You know its mass and therefore weight (remember weight being a force). Also, you have acceleration "a" and the mass, so again, a force.

Once you solve part (a), think of part (b) in a similar fashion.
 
  • #3
I just don't know what to do
 
  • #4
angiep410 said:
I just don't know what to do

List the relevant equations, and use the hints you got in Post #2. Then start working through the problem, and show your work here.
 
  • #5
a= f/m, but I don't know what the force is
the mass is 65 kg so the force is 637.4 N so when I divide 637.4/65 I get 9.8 (gravity), but that really hasn't gotten me anywhere, has it?
I don't know the equation for tension. We haven't learned it yet but the assignment is due Monday.
 
  • #6
angiep410 said:
a= f/m, but I don't know what the force is
the mass is 65 kg so the force is 637.4 N so when I divide 637.4/65 I get 9.8 (gravity), but that really hasn't gotten me anywhere, has it?
I don't know the equation for tension. We haven't learned it yet but the assignment is due Monday.

On these kind of questions, you start by drawing a Free Body Diagram (FBD) of the object in question. The FBD needs to show all the forces acting on the object.

What forces do you show on your FBD? What directions do the forces point on the object?

And then you use what you listed, F=ma. You know the acceleration vector, and that acceleration is due to the sum of the forces. In fact, the better way to write F=ma for this is:

[tex]\Sigma F = m a[/tex]
 
  • #7
In my FBD, I have gravity pointing down, the force of the ground pointing up, (the force of the ground and the force of gravity are equal), friction pointing to the left, and acceleration pointing to the right (acceleration is larger since the van has constant acceleration.
 
  • #8
angiep410 said:
In my FBD, I have gravity pointing down, the force of the ground pointing up, (the force of the ground and the force of gravity are equal), friction pointing to the left, and acceleration pointing to the right (acceleration is larger since the van has constant acceleration.

Since the weight is suspended by a string, the ground does not directly exert a force on the weight. You are correct that the weight force on the object points down. Which way does the force of tension in the string point for the object's FBD? And then can you list all the forces on the object, and write out the equation for:

[tex]\Sigma F = m a[/tex]
 
  • #9
BTW, when you sum the forces, you sum the components. That is, you set the sum of all x-direction forces to the x-direction m*a, and you set the sum of the y-direction forces to the y-direction m*a.
 
  • #10
The force of tension would point up because that's where the tension is coming from. (I think). I understand the point of FBD's, but if I don't know the values then how does it help, in this case?
 
  • #11
angiep410 said:
The force of tension would point up because that's where the tension is coming from. (I think). I understand the point of FBD's, but if I don't know the values then how does it help, in this case?

Remember that the tension can only act along the string, so not exactly straight up, but at the same angle from straight up as listed in the problem.

And Drawing a FBD is helpful because it shows you which components are equal to each other. You can set your unknowns equal to your knowns and solve them. It will help you keep track of the problem and visualize it.
 
  • #12
Yes, I know that is why we use FBD, but in this case all I know is the mass and an angle. I don't really understand what I'm setting equal to each other.
 
  • #13
angiep410 said:
Yes, I know that is why we use FBD, but in this case all I know is the mass and an angle. I don't really understand what I'm setting equal to each other.

Show us your FBD. Can you scan it or sketch it and attach it to your post?

You will break each force into its x and y components, and set the sum in each dimension equal to m*a.

Is the mass accelerating in the vertical (y) direction? How about in the x direction? What does that say about the sum of the forces in the y direction? How about in the x direction?
 
  • #14
I don't have a scanner or anything. I guess no one can help me. I understand and have done everything you guys are saying, and I appreciate the help. I was just looking for a more direct approach to the problem instead of everyone telling me what I have already done.
 
  • #15
angiep410 said:
I don't have a scanner or anything. I guess no one can help me. I understand and have done everything you guys are saying, and I appreciate the help. I was just looking for a more direct approach to the problem instead of everyone telling me what I have already done.

There's not really a more direct approach. You pretty much always start with a FBD, and then sum force components in 2 or 3 dimensions.

Do you have Paint on your PC? You can do simple sketches in Paint, and save them as JPG files to attach to your posts. You will find the ability to post sketches in your PF threads to be very helpful...
 

1. How do I find the acceleration given mass and an angle?

To find the acceleration, you will need to use the formula a = gsinθ, where "a" represents acceleration, "g" represents the acceleration due to gravity (usually 9.8 m/s²), and θ represents the angle of the inclined plane.

2. Can I find the acceleration without knowing the mass?

No, you will need to know the mass of the object in order to calculate the acceleration. The formula for acceleration (a = gsinθ) includes the mass of the object, so it is necessary to have this information in order to solve for acceleration.

3. What is the equation for finding tension in a string or rope?

The formula for tension is T = mgcosθ, where "T" represents tension, "m" represents the mass of the object, "g" represents the acceleration due to gravity, and θ represents the angle of the incline. This formula assumes that the string or rope is in a state of equilibrium, meaning that the net force on the object is zero.

4. Do I need to convert the angle to radians when using these formulas?

Yes, the angle must be in radians in order to use the formulas for acceleration and tension. To convert from degrees to radians, you can use the formula θ(rad) = θ(deg) x π / 180.

5. What if the object is not on an inclined plane?

If the object is not on an inclined plane, then the formulas for acceleration and tension will not apply. These formulas are specifically for objects on an inclined plane, where the force of gravity is acting at an angle. For objects in other scenarios, different equations will need to be used to calculate these values.

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