A uniform 300-kg beam,6.00 m long,is freely pivoted at P,as shown in the figure.The beam is supported in a horizontal position by a light strut,5.00 m long,which is freely pivoted at Q and is loosely pinned to the beam at R.A load of mass is suspended from the end of the beam at S.A maximum compression of 23,000 N in the strut is permitted,due to safety.Under maximum load,find the magnitude of the x component of the force exerted on the beam by the pivot at P.

Question

Quizplus · Accepted Answer

To solve this problem, we need to apply the principle of moments, which states that the sum of clockwise moments about any point must be equal to the sum of anticlockwise moments about the same point. We will choose point P as the pivot and take moments about P.

Let the weight of the beam be W1, the weight of the load be W2, and the tension in the strut be T. Since the beam is uniform, its weight acts at its midpoint, which is 3.00 m from P. The load is at a distance of 6.00 m from P.

Taking moments about P, we get:

T × 5.00 m - W1 × 3.00 m - W2 × 6.00 m = 0

Solving for T, we get:

T = (W1 × 3.00 m + W2 × 6.00 m) / 5.00 m

Now, we need to find the maximum value of T that satisfies the given constraint that the compression in the strut should not exceed 23,000 N. We know that:

T / sinθ = 23,000 N

where θ is the angle between the strut and the horizontal. Using trigonometry, we can find that:

sinθ = 3.00 m / 5.10 m

cosθ = 4.10 m / 5.10 m

Substituting these values, we get:

T = 23,000 N × sinθ = 23,000 N × 3.00 m / 5.10 m = 13,529 N

Therefore, the maximum value of T is 13,529 N, which is less than the allowed maximum compression.

Now, we can find the horizontal component of the force exerted on the beam by the pivot at P. This is simply the sum of horizontal components of T and W1. Since these forces act perpendicular to each other, we can use the Pythagorean theorem to find their resultant:

F = √(T² + W1²)

Substituting the values, we get:

F = √(13,529 N² + 8,823 N²) = 16,000 N

Therefore, the magnitude of the x component of the force exerted on the beam by the pivot at P is 16,000 N. The correct answer is A.

A Uniform 300-Kg Beam,6