(896) Problem 7: A fence post of mass m 6 kg supports a fence with three lengths
ID: 2041397 • Letter: #
Question
(896) Problem 7: A fence post of mass m 6 kg supports a fence with three lengths of barbed wire. The bottom wire is a distance d-0.35 m from the ground and each wire is a distance 0.35 m above the previous one. Each of these three wires has the tension on it. For this problem, assume that the force exerted by these wires is purcly horizontal. An additional guy wire is used to kcep the pole upright with a tension of Fa 510 N. The guy wire attaches to the post at a point h-0.89 m above the ground and makes an angle of ? = 35 with respect to the horizontal Otheexpertta.com 25% Part (a) Choose the correct free body diagram Grade Summary Deductions Potential Late work % 50% Late Potential 50% 0% Submissions Attempts remaining: 5 Fx 0% per attempt) detailed view ng mg Submit Hint I give up! Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback 25% Part (b) Write an expression for the tension in any one of the three fence wires 25% Part (c) Find the tension, in newtons, in any one of the three fence wires ? 25% Part (d) What is the normal force that the ground exerts upward on the post?Explanation / Answer
a) 3rd option(upper right figure) is the correct answer.
b) As the post is in equilibrium net force and net torque acting on it must be zero.
Apply net torque about bottom of the post = 0
T*d*sin(90) + T*2*d*sin(90) + T*3*d*sin(90) - FA*h*sin(90-theta) = 0
6*T*d - FA*h*sin(90-35) = 0
T = FA*h*sin(55)/(6*d) <<<<<<<<<<<------------------Answer
c) T = 510*0.89*sin(55)/(6*0.35)
= 177 N <<<<<<<<<<<------------------Answer
d) now apply, Fnety = 0
Fn - FA*sin(theta) - m*g = 0
Fn = FA*sin(theta) + m*g
= 510*sin(35) + 6*9.8
= 351 N <<<<<<<<<<<------------------Answer
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