How to find acceleration and tension given mass and an angle

AI Thread Summary
To find the acceleration of a van and the tension in a string suspending a mass, start by analyzing the forces acting on the mass using a Free Body Diagram (FBD). The weight of the mass (637.4 N) acts downward, while tension acts at an angle due to the van's acceleration. The problem requires breaking down these forces into their x and y components and applying Newton's second law (ΣF = ma) for each direction. The mass does not accelerate vertically, indicating that the sum of forces in the y-direction equals zero, while the x-direction must account for the van's acceleration. A clear understanding of these components is essential for solving the problem effectively.
angiep410
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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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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.
 
I just don't know what to do
 
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.
 
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.
 
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:

\Sigma F = m a
 
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.
 
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:

\Sigma F = m a
 
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...
 
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