At a construction site, a pallet of bricks is to be suspended by attaching a rop

Question

At a construction site, a pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown in the figure. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.777.(Figure 1)

Part A

What is the weight of the heaviest pallet of bricks that can be supported this way? Start with appropriate free-body diagrams.

Express your answer in pounds to three significant figures.

311

Correct

Part B

What is the friction force on the upper crate under the conditions given in Part A?

Express your answer in newtons to three significant figures.

Figure 1 of 1

311

Explanation / Answer

Total mass on the roof is m = 150 + 250 = 400 lbs

Co-efficeint of friction u = 0.777

Frictional force towards left = 400 x 0.777 x g

Tension in the upper part of the rope T = 400 x 0.777, tension in the rope is equal to the frictional force to be at rest

If m is the mass of bricks the gravitational force acting on the bricks

mg - this equal to the tension in the rope

Now we have mg = 400 x 0.777 x g

                       m = 311 lbs

Part B is not very clear, it is assumed that we have only upper crate of weight 150 lbs on the roof

if u is the co-efficient of friction then frictional force u 0.777

       = 150 x 0.777 g = 116 g

=116 x 0.44 x 9.81, we take 0.44 as conversion from lb to kg and 9.81 as g

=500.70 N

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