Solving a Physics Problem: Finding Pole Mass, Forces & Distance

[PhilCam]

Homework Statement

At the end of the semester Peter and Paul are moving out of the dorms. The are holding a 6 kg box hung from a homogeneous pole of unknown mass. The position from which the 6 kg box is hung is also unknown. The length of the pole is 1.4 m. As they carry the box, the pole is horizontal. At the left end of the, Peter applies a force of 67.9 N making an angle of 60 degrees with the horizontal. At the right end, Paul applies an unknown force making a 30 degree angle with the horizontal. The system is in static equilibrium.

A) The x-component of the force applied by Peter is?
B)The magnitude of the force applied by Paul is?
C)The unknown mass of the pole is?
D) The distance to the left end of the pole form the point where the 6 kg mass is hung is?


I have figured out A by 67.9 x Cos 60 = -33.95.

For part B, I assumed that the magnitude of Paul would have to equal the magnitude of Peter's force but I have not been able to come up witht he answer. Thanks.

 

[ideasrule]

Start by drawing a free-body diagram of the pole. Remember that in static equilibrium, all forces and all torques must sum to 0.

 

[kerimek]

I would approach this problem by first drawing a free-body diagram of the system and labeling all the known and unknown forces. Then, using the principles of static equilibrium, I would set up equations to solve for the unknowns.

For part A, the x-component of Peter's force can be calculated using the equation Fx = Fcosθ, where θ is the angle between the force and the horizontal. In this case, it would be -33.95 N.

For part B, the magnitude of Paul's force can be found by setting up an equation for the sum of forces in the y-direction, since the system is in static equilibrium. This would be: Fy = Fsinθ + 6kg x g = 0, where g is the acceleration due to gravity. Solving for F, we get F = -29.4 N.

For part C, we can use the equation for torque, τ = r x F, where r is the distance from the point where the force is applied to the point where the mass is hung. Since the system is in static equilibrium, the torque must be equal to zero. Therefore, we can set up the following equation: r x 67.9cos60 - (1.4-r) x 29.4cos30 = 0. Solving for r, we get r = 1.18 m. Then, using the equation for torque again, we can solve for the unknown mass of the pole: 1.18 x 67.9cos60 = (1.4-1.18) x M, where M is the unknown mass. Solving for M, we get M = 7.88 kg.

For part D, the distance to the left end of the pole from the point where the 6 kg mass is hung can be found by subtracting the distance r from the total length of the pole, 1.4 m. Therefore, the distance would be 0.22 m.

In conclusion, by approaching this problem systematically and using the principles of physics, we were able to solve for all the unknown variables and find the answers to the given questions.