Suggestion
Draw a FBD of the man and the cart together. This FBD will have three tension forces acting on it: two at the pulley on the cart and one on the man. Newton's 2nd Law will give the acceleration of the cart and man together.
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7 comments:
is the tension equal to the 60lb?
I am having trouble identifing the tension as well. Would it be 2T + P = 0, but if you do that than it will cancel each other out. Any suggestions?
I am having that same trouble just getting started too. How are there two tension forces on the cart. Wouldn't there just be one since there's only one actual string connected to the cart from the pulley?
-rjaneshe- Both of the tensions from the pulley attached to the cart act at the same point.
-deborah & courtney- The tension in the rope is equal at all points in the rope. This means the tension in all three sections equals the force from the man on the rope. All of these tensions appose the cart rolling down the incline.
-courtney- The cart is on an incline so the sum of the tensions cannot equal zero. The sum of the tensions is equal to a non-zero force. If the sum of the tensions equaled zero, the cart would roll down the incline. Also, this situation is not static. The sum of the forces along the incline will not equal zero. Their sum will equal mass*acceleration, mass equals weight of the cart/man divided by gravity, acceleration is what you are trying to solve for.
so is there a x and y component of acceleration
If you put your x axis parallel to the incline then you only have motion in that direction. The cart will only move in the x direction and you won't have any y components to worry about.
Let me jump in for a couple comments.
First, great comments by Scott and mark garety.
Second, if you are still wondering about how to start out with your FBD, please look back at the figure that I just added to the original post.
A "free body diagram" means that you cut your system away from its surroundings leaving a "free body". When you cut away the surroundings from the cart + man, you have a normal force at each wheel, a tension force at each place that a cut goes through the cable and a weight force for the cart and man. As you can see in the figure, this cut exposes three tension forces. As Scott explained, the tension is constant throughout the cable and is equal to the force of the man on the cable (60lb).
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