A 20 kg sign is pulled by a horizontal force such that the single rope holding the sign makes an angle of 21 degrees with the vertical. Assuming the sign is motionless, find magnitude of the tension in the rope.
Well we know that our whole system should have a net torque of 0. So the Torque caused by tension needs to cancel with the torque of both the beam center of mass and the sign's torques. so TL-1/2Lmgsin(21)-Lmgsin(21)=0 the lengths cancel leaving T - 1/2mgsin(21)-mgsin(21)= 0
To solve this problem, we can use the concept of resolving forces into horizontal and vertical components.
Step 1: Draw a diagram of the situation.
We have a sign being pulled by a horizontal force, and a rope making an angle of 21 degrees with the vertical.
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/ |
F/ |
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/θ |
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Sign
Step 2: Identify the forces acting on the sign.
There are two forces acting on the sign: the tension force (T) in the rope and the gravitational force (mg) pulling the sign downwards.
Step 3: Resolve the forces into horizontal and vertical components.
The tension force in the rope can be broken down into two components: T_x in the horizontal direction and T_y in the vertical direction.
T_x = T * cos(θ)
T_y = T * sin(θ)
Step 4: Equate the vertical forces to determine the tension in the rope.
Since the sign is motionless, the vertical forces must balance each other.
T_y = mg
Step 5: Substitute the values and solve for T.
T * sin(21) = (20 kg) * (9.8 m/s^2)
T = (20 kg) * (9.8 m/s^2) / sin(21)
T ≈ 97.8 N
Therefore, the magnitude of the tension in the rope is approximately 97.8 N.