During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 253 N on the wire. The left section of the wire makes an angle of 15.4° relative to the horizontal and sustains a tension of 459 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Question

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 253 N on the wire. The left section of the wire makes an angle of 15.4° relative to the horizontal and sustains a tension of 459 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

bob · Accepted Answer

how did you get the direction in degrees? what do you do with the Cot?

Henry · Answer

The system is in equilibrium:
T1*sin15.4 + T2*sin A = +253.
459*sin15.4 + T2*sin A = 253,
121.9 + T2*sin A = 253.
T2 = (253-121.9)/sin A = 131.1/sin A.

-T1*Cos15.4 + T2*Cos A = 0,
-459*Cos15.4 + T2*Cos A = 0.
T2*Cos A = 459*Cos15.4 = 442.5,
(131.1/sin A)Cos A = 442.5,
131.1*Cos A/sin A = 442.5,
131.1*Cot A = 442.5, A = 16.5 Deg. = Direction.

T2 = 131.1/sin16.5 = 461.6 N.