a 10kg flowerpot is suspended from the end of a horizontal strut by a cable attached at 30° above the horizontal. If the strut h
as no mass what is tension in the cable
1 answer:
T = tension force in the cable
m = mass of the flowerpot = 10 kg
θ = angle cable makes with the horizontal = 30 deg
there are horizontal and vertical components of tension force . The vertical component of tension force in cable is in opposite direction of the weight of the flowerpot. hence vertical component of tension balances the weight of flowerpot. so we can write
T Sinθ = mg
inserting the above value
T Sin30 = 10 x 9.8
T(0.5) = 98
T = 98/0.5
T = 196 N
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Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then
From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
