As shown in the figure above, a thin horizontal bar AB of negligible weight and
ID: 1400517 • Letter: A
Question
As shown in the figure above, a thin horizontal bar AB of negligible weight and length L = 6.4 m is pinned to a vertical wall at A and supported at B by a thin wire BC that makes an angle ? = 32° with the horizontal. A weight W = 300 N can be moved anywhere along the bar; its position is defined by the distance x from the wall to its center of mass.
As a function of x, find the tension in the wire, and the horizontal and vertical components of the force exerted on the bar by the pin at A.
If x = 2.98 m, what is the tension?
If x = 2.98 m, what is the magnitude of the horizontal component of the force exerted by the pin?
If x = 2.98 m, what is the magnitude of the vertical component of the force exerted by the pin?
Explanation / Answer
weight force of block will act on rod in downwards direction.
if we find torque of rod about hinge point, then torque due force of pin will be zero as r = 0 for these force
torque = r X F = rFsin@
rod is equilibrium, so balancing torque about hinge (pin) point.
x*W - L* T*sin3@ = 0
T = Wx / Lsin@ .......Ans
300x - 6.4Tsin32 = 0
T = 88.46x .................Ans
When x = 2.98 m
So, T = 88.46 x 2.98 = 263.60 N ..........Ans
balancing forces in horizontal direction,
Fx - Tcos32 = 0
Fx = 263.60cos32 = 223.55 N
balancing forces in vertical direction,
W - Fy - Tsin32 = 0
Fy = 300 - 263.60sin32 = 160.31 N .....Anss
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