Slightly better photo
First step is to draw a good diagram and label all the forces. You should label tension in the cable, weight of girder, friction and reaction force. Then do as joined says and take moments about the contact point (which handily eliminates friction and reaction from the equation leaving you with just components of weight and tension.
Then for part b, just resolve forces horizontally and vertically to get each component of the 'support force' (I've never seen this term before, which exam board is this?). Hope that helps
Then for part b, just resolve forces horizontally and vertically to get each component of the 'support force' (I've never seen this term before, which exam board is this?). Hope that helps
I tried the question, but I noticed something weird.
First I note that there are four forces on the girder: gravity, tension, the normal, and friction. The girder is not moving, so the total of these forces is zero.
Then I divide the gravity into two parts: the force perpendicular to the girder, and hence parallel to the tension, and the part parallel to the girder.
The part perpendicular to the girder must be balanced by the tension. Therefore W sin ø = T. Or W cos ø = T. It doesn't really matter which angle I chose here. Say W cos ø where cos ø = 3/5.
Then comes the weird part. Since the perpendicular weight is already balanced, the sum of the normal and the friction must balance the parallel weight, which is W sin ø. The horizontal component of W sin ø, W sin ø cos ø is equal to the friction. Since W = 15000 N, f = 7200 N which is a long way off from the book's answer.
Anyway, using the same method gives the normal to be 5400 N, again a long way off the book's answer.
And I am pretty sure of my answers, because I was able to get them in two different ways: force analysis and virtual work.
So I checked the book's answers. According to the book the normal is 10200 N and friction is 3.6 N. Only the horizontal component of tension, T cos ø and friction are acting horizontally on the girder, hence f = T cos ø. The vertical forces on the girder are W, N and T sin ø, which means W = N + T sin ø. Substituting for T from f = T cos ø to solve for W gives
W = N + f tan ø, where tan ø = 4/3
Using this I can check the answers: I know that W = 15000. If I substitute the values of N and f from the book, I get something round 10.2N, which can't be right. If I substitute the values I worked out, 5400 and 7200, I get 15000. Besides, it makes sense that the friction is greater than N, since I'm dragging iron on the ground, so the coefficient of friction can't be less than 1. And even if it was, it does not make sense that it is smaller than 0.001, which the book seems to say;
f = uN.
So what did I miss here?
Thank you.
First I note that there are four forces on the girder: gravity, tension, the normal, and friction. The girder is not moving, so the total of these forces is zero.
Then I divide the gravity into two parts: the force perpendicular to the girder, and hence parallel to the tension, and the part parallel to the girder.
The part perpendicular to the girder must be balanced by the tension. Therefore W sin ø = T. Or W cos ø = T. It doesn't really matter which angle I chose here. Say W cos ø where cos ø = 3/5.
Then comes the weird part. Since the perpendicular weight is already balanced, the sum of the normal and the friction must balance the parallel weight, which is W sin ø. The horizontal component of W sin ø, W sin ø cos ø is equal to the friction. Since W = 15000 N, f = 7200 N which is a long way off from the book's answer.
Anyway, using the same method gives the normal to be 5400 N, again a long way off the book's answer.
And I am pretty sure of my answers, because I was able to get them in two different ways: force analysis and virtual work.
So I checked the book's answers. According to the book the normal is 10200 N and friction is 3.6 N. Only the horizontal component of tension, T cos ø and friction are acting horizontally on the girder, hence f = T cos ø. The vertical forces on the girder are W, N and T sin ø, which means W = N + T sin ø. Substituting for T from f = T cos ø to solve for W gives
W = N + f tan ø, where tan ø = 4/3
Using this I can check the answers: I know that W = 15000. If I substitute the values of N and f from the book, I get something round 10.2N, which can't be right. If I substitute the values I worked out, 5400 and 7200, I get 15000. Besides, it makes sense that the friction is greater than N, since I'm dragging iron on the ground, so the coefficient of friction can't be less than 1. And even if it was, it does not make sense that it is smaller than 0.001, which the book seems to say;
f = uN.
So what did I miss here?
Thank you.
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