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A 10kg block is suspended by two ropes. Each rope makes an angle of 45 degrees to the horizontal. What is the magnitude of the tension force in each rope? g= 10 m/s2 W= 100 N
A 10kg block is suspended by two ropes. Each rope makes an angle of 45 degrees to the horizontal. What is the magnitude of the tension force in each rope? g= 10 m/s2 W= 100 N
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Answer:
10/2/100, 1200
Explanation:
10,20 160
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Answer:
Assuming the roped to be weightless, as the ropes are symmetrically supporting the block, each had the same tension say T Newtons.
Vertical components ot T in each rope,
Th = T* Cos 45°=T/√2
Sum of the vertical components in the two ropes
=2Th=2T/√2= T√2 Newtons.
Vertical downward force = weight of the block
= 10*g= 10*9.8= 98 N.
At the block is stable, vertical upward bl and downward forces are equal ie
T√2 = 98 or T = 98/√2= 69.3 N.
Thus, the tension in each rope is 69.3 N.
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