The figure shows a 100 kg plank supported at its right end by a 7.0-mm-diameter rope with a tensile strength of 6.0*10 ^7 N/M^2. How far along the plank, measured from the pivot, can the center of gravity of an 800 kg piece of heavy machinery be placed before the rope snaps?
Here is a link to the picture: http://www.chegg.com/homework-help/questions-and-answers/figure-shows-100-rm-kg-plank-supported-right-end-70-rm-mm-diameter-rope-tensile-strength-6-q2784402
I'm pretty sure this is a torque problem, but im not sure how to start it exactly. Any help would be great!
Here is a link to the picture: http://www.chegg.com/homework-help/questions-and-answers/figure-shows-100-rm-kg-plank-supported-right-end-70-rm-mm-diameter-rope-tensile-strength-6-q2784402
I'm pretty sure this is a torque problem, but im not sure how to start it exactly. Any help would be great!
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Yes, this is a simple torque/moment problem.
We know we'll want the rope's ultimate tensile strength, so let's solve for that. We have been told that the strength is 6.0*10^7 Newtons per square meter of rope, and that the rope itself is 7.0mm in diameter. From there, we can calculate the rope's sectional area (assume it's a circle) and then the tensile strength of the rope... about 2310 Newtons is the number I come up with.
With this information, we can look now at our free body diagram of our plank. On it, we have:
a pivot reaction (providing force up)
the weight of the board (981 Newtons down at the center of the board)
2310 Newton tension from the rope going up at the far end of the board
and the 7848N force coming from the machinery.
It should be noted that on the picture I am viewing from your link, the board length is 3.5m. You'll have to correct the math for any variations yourself.
For now, let us assume that the pivot provides all the force up needed to keep the board from falling and save ourselves the trouble of writing that equation. Let us simply assume that we want to prevent the board from tipping, and to do that, we set up an equation for the moment about the pivot (since we didn't solve for the reaction there, doing this allows us to ignore it).
The machinery provides a moment of 7848x Nm clockwise (call it negative).
The plank's weight provides a moment of 1715 Nm clockwise (call it negative).
The rope provides a moment of 8085 Nm counter-clockwise (call it positive).
We want the total moment about the pivot to equal zero, so:
0 = 8085 - 1715 - 7848x
-6370Nm = -7848x
x = 0.812 m
Thus, the distance the machinery can move from the pivot is 0.812m, according to the picture I see.
We know we'll want the rope's ultimate tensile strength, so let's solve for that. We have been told that the strength is 6.0*10^7 Newtons per square meter of rope, and that the rope itself is 7.0mm in diameter. From there, we can calculate the rope's sectional area (assume it's a circle) and then the tensile strength of the rope... about 2310 Newtons is the number I come up with.
With this information, we can look now at our free body diagram of our plank. On it, we have:
a pivot reaction (providing force up)
the weight of the board (981 Newtons down at the center of the board)
2310 Newton tension from the rope going up at the far end of the board
and the 7848N force coming from the machinery.
It should be noted that on the picture I am viewing from your link, the board length is 3.5m. You'll have to correct the math for any variations yourself.
For now, let us assume that the pivot provides all the force up needed to keep the board from falling and save ourselves the trouble of writing that equation. Let us simply assume that we want to prevent the board from tipping, and to do that, we set up an equation for the moment about the pivot (since we didn't solve for the reaction there, doing this allows us to ignore it).
The machinery provides a moment of 7848x Nm clockwise (call it negative).
The plank's weight provides a moment of 1715 Nm clockwise (call it negative).
The rope provides a moment of 8085 Nm counter-clockwise (call it positive).
We want the total moment about the pivot to equal zero, so:
0 = 8085 - 1715 - 7848x
-6370Nm = -7848x
x = 0.812 m
Thus, the distance the machinery can move from the pivot is 0.812m, according to the picture I see.
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its calculation