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A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg weight is suspended at 30 cm from left end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.) (1) 75 N, 49 N (2) 55 N, 43 N (3) 11 N, 25 N (4) 7 N, 24 N

A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg weight is suspended at 30 cm from left end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)
(1) 75 N, 49 N
(2) 55 N, 43 N
(3) 11 N, 25 N
(4) 7 N, 24 N

Grade:12th pass

2 Answers

Saurabh Kumar
askIITians Faculty 2411 Points
6 years ago
245-864_999.PNG
ankit singh
askIITians Faculty 614 Points
one year ago
here we take g = 10 m/s2
We shall consider the situations in which the rod is both translationally  and rotationally stable. Two types of forces act on the rod, the weight and the force normal reaction.
Lets have a look at the appropriate figure
 
 
Two knife edges are placed at the two ends of the rod at positions A and B. They both exert normal reactions on the rod given as R and r respectively (we don;t know which one is greater). They are directed vertically upwards.
A body of weight of mass (M = 6 kg) is W and is placed about 30 cm from one end (the left one in this case). The center of gravity of the body (w) has mass (m) 4 kg, it is the point where all of its mass is supposed to be concentrated and is denoted as C. Both these points exert the force due to their weight and is directed downwards.
Now, for the body to have translational stability, the forces due to normal reaction should balance the force due to the weights.
Thus
R + r = W + w   (1)
here W = Mg =  6 X 10 N
and  w =  mg = 4 x 10 N
so,   R + r = 100 N  (2)
 
Now for the body to be rotationally stable (so, it experiences no rotation), all the opposite forces should balance out.
It is easier to consider the rotation about the center of gravity (C), we can ignore this force.
Due to their directions, the reaction on the knife edge A (R) would tend to impart the rod with clockwise motion (upwards and right). However, the reaction r and weight W would act in the anti-clockwise directions (it is because the weight is placed on the left of the center of gravity, we could ahve either way). So, here themoment causes rotation, thus
the moment on knife edge A would be = force of reaction X distance from the center of gravity (AC)
similarly, we can calculate moment for knife edge at B and due to the mass M.
thus, the equilibrium equation would become
 
(R X AC)  = (W X CM) + (r X BC)   (3)
 
converting the distances to meters and substituting the values, we would get
 
 0.25R = 0.05X60 + 0.25r
or   R - r = 12 N  (4)
 
Now, by using equations (2) and (4), we can calculate the values of R and r.
finally, we get
 
R = 56 N and
r = 44 N

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