7. A 100-kg nonuniform boom that is 6.0 m long is loosely...

7. A 100-kg nonuniform boom that is 6.0 m long is loosely pinned at the pivot at P. A 600-kg concrete block is suspended from the end of the boom at A, as shown in the figure. The boom forms a 30° angle above the horizontal, and is supported by a 4.0-m cable between points D and B. Point B is 4.0 m from P, and point D is 4.0 m above P. The center of mass of the boom is at point C, which is 2.0 m from P. Assume that all the quantities shown in the figure are accurate to two significant figures. What is the tension in the cable connected between points B and D? A) 9,300 N B) 8,400 N C) 8,100 N D) 7,500 N E) 6,900 N

Transcript

00:01 Hello student, in this question we are given about this arrangement which is shown in this diagram that a rod which is of non -uniform density has having a mass of 100 kilograms is being pivot at a point p which is at this point p is pivot and we are given that at another end of the road from this pivot at point a a concrete block of mass 600 kilograms is hanged and we are given that at a point 4 meters from the rod that is from the pivot point at point b a string is attached to the wall which is on which the pivot is also attached and we are given that at a distance of 2 meters from the pivot point that is at point c lies the center of mass of the draw so we are given that the system is in equilibrium and we have to calculate the tension in the string b tension in the string is b question mark or we have to calculate the tension in the string bd so let us first assume the tension which would be acting in this string as t we are given that the string length is of 4 centimeters the distance between p and b is also 4 centimeters and p and b and d and p is also 4 centimeters so all the units are in meters so this is an equilateral triangle so we can see that the angle between this angle would be equal to 60 degrees which means if we try to resolve the components of tension along the vertical and horizontal line so we can do that easy as we know that this angle we know that the center of mass is acting at this point c so the mass of the whole mass of the road can be assumed at this point acting in the downward direction mass of the road can be noted by a small letter m in the mass of the concrete road can be noted weight w mg would be acting at point c which is a 2 cm from p 2 meters from p point and now we have to calculate the tension in the string so first of all as a system is in equilibrium we can balance the moment in the road along about point p moment about point so as we know that and is acting in this direction and this angle this angle is given in the quotient at 30 dg so this angle would be equal to 60 degrees so if we try to resolve the component of mg along the plane and the plane perpendicular to the rod or m g would be equal to this angle would be this component of the weight of the block would be equal to mg sine of 60 degrees if we try to resolve the tension along the plane one is the perpendicular parallel plane and another is the perpendicular plane to the rod so the component of tension along the perpendicular component would be equal to t sine of 60 degrees and if we try to resolve the component of weight which would be acting in this direction so perpendicular would be acting in this direction and this angle again would be equal to this angle would again be equal to 30 degrees which means this angle would also be equal to 60 degrees.

03:42 So taking the component of weight perpendicular to the rod, it is coming out to the w sign of 60 degrees.

03:50 So if we try to take the moment about point p, this mg moment and this weight moment is trying to rotate the rod in the clockwise direction but this component of tension which is acting in upward direction would be applying.

04:07 An anti -clockwise moment about point p as they are balancing each other it is given the question the object the system is in equilibrium so we can just balance the moment about point p which would be equal to the moment about the first of all the clockwise torque would be equal to the anti -clockwise torque which is equal to mg sine of 60 degrees that is the moment due to the mass of the road and and multiplied by the distance from the point that is equal to 2 meters plus the another clockwise torque is the torque due to the weight which is equal to w sine of 60 degrees multiplied by the distance which is 6 meters and this moment that is the clockwise moment it would be equal to the anti -clockwise moment which is equal to tension sine of 60 degrees multiplied by the distance which is given in the time...

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