A uniform drawbridge must be held at a 335 angle above the...

A uniform drawbridge must be held at a 33.5 ∘∘ angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 44000 NN, is 10.5 mm long, and pivots about a hinge at its lower end. A cable is connected 3.80 mm from the hinge, as measured along the bridge, and pulls horizontally on the bridge to hold it in place. A. What is the tension in the cable? B. Find the magnitude of the force the hinge exerts on the bridge? C. Find the direction of the force the hinge exerts on the bridge? D. If the cable suddenly breaks, what is the initial angular acceleration of the bridge?

Transcript

00:01 For the given problem if we consider that this is force of y, this is force of x, this is f x, this is f x, and where we have this to be, from here, this is the tension t, this is point a, and from the center, there is a weight downwards which is acting, and this angle is 53 .5 degree, and over here there is point b, and this is rotating in this manner.

00:33 We can say that if we suppose this is point o so we can say from here the value of oa that is given in the problem is 3 .80 meter and the value of l length that is equals to o b total length is given as 10 .5 meter so at equilibrium if we conserve the moment to be equal to zero so we can say that moment due to weight of bridge moment due to weight of bridge will become m1 will be force on weight multiplied to l upon 2 cos of theta will be equals to m g l upon 2 cos theta thus the value of m1 will come out to be 44 hundred sorry 44 ,000 multiply to 10 .5 upon 2 cos 3 3 .5 .5 and thus this value will come out to be 192627 .6 .9 .9 newton.

01:47 Now in the case, that is moment due to, that is other moment or moment due to cable, will be m2 will be equal to minus t times oa sine theta, will also be equal to minus t multiplied to 3 .80 sine 3 .30 so if we find out the value of m1 plus m2 to be equal to 0, net moment will be equal to 0, this will be minus t multiplied to 3 .80, sine 33 .5 plus 192627 .67 .6 will be equal to 0.

02:31 Thus from here the value of tension will come out to be 91893 newton.

02:39 This is the tension in the cable.

02:41 So if we find the x component, the x component will be equal to the tension that is also equal to 9189 .4 .3 newton.

02:52 Whereas the y component will be equal to f of w weight, that is 44080 newton.

03:01 So the resultant will be square root of 918943 square plus 440...

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