A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn
itude of the tension in the wire if the angle between the cable and the horizontal is θ = 47°.
1 answer:
Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
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Answer:
(i) 7.2 feet per minute.
(ii) No, the rate would be different.
(iii) The rate would be always positive.
(iv) the resultant change would be constant.
(v) 0 feet per min
Explanation:
Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,
By making the diagram of this situation,
Applying Pythagoras theorem,
Differentiating with respect to t ( time ),
( l = 26 feet = constant )
We have,
(i) From equation (1),
From equation (X),
(ii) From equation (X),
Thus, for different value of x the value of would be different.
(iii) Since, distance = Positive number,
So, the value of y will always a positive number.
Thus, from equation (X),
The rate would always be a positive.
(iv) The length of the ladder is constant, so, the resultant change would be constant.
i.e. x = increases ⇒ y = decreases
y = decreases ⇒ y = increases
(v) if ladder hit the ground x = 0,
So, from equation (X),
