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PHYS 170 Lecture 6: week 6 notes

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School
UBC
Department
Physics
Course
PHYS 170
Professor
Mcmillan Malcolm

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Document Summary

Impending tipping (page 390) page 2: figure 8-1 (d) Figure 8-1 (d) shows a uniform block of weight. P acts on the block. rough horizontal surface. The surface exerts a resultant normal force resultant friction force. N acts a distance x to the right of the line of action of. Wx = ph: criterion for no tipping. The block will not tip if x = The block is on the verge of tipping when x = Impending sliding (page 390) page 2: figure 8-3. Figure 8-3 shows friction force f as a function of applied force p . It is assumed that tipping does not occur. Impending sliding (page 390) page 3: no motion. The block is in equilibrium when f = p and f ! S is the coefficient of static friction between the block and the surface. The block is on the verge of sliding when f = f s . This situation is called impending sliding: motion.

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Related Questions

In a laboratory experiment on friction, a 135-N block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the block begins to move and continues to increase thereafter. Figure 5.52 shows a graph of the friction force on this block as a function of the pull.
(a) Identify the regions of the graph where static and kinetic friction occurs.
(b) Find the coefficients of static and kinetic friction between the block and the table.
(c) Why does the graph slant upward in the first part but then level out?
(d) What would the graph look like if a 135-N brick were placed on the box and what would be the coefficients of friction be in that case?

 

A person is pulling a block of mass m w/ a force equal to itsweight directed 30 degrees above the horizontal plane across arough surface generating a friction f on the block. If the personis now pushing downward on the block with the same force 30 degreesabove the horizontal plane across the same rough surface what isthe friction on the block? Answer is 3f

Explanation states F=ukN where N is normal force acting on block. Iunderstand that however it then states when force is applied 30degrees above horizontal N=mg-mgsin30? How is that the normal forceI thought the normal was mgcos30? It the states when force isapplied 30 degrees below horizontal N=mg+mgsin30? Solving for N andplugging it into F=ukN and solving for uk gives 3f.

A horizontal force F1=75 N and a force F2=16.5 N, acting at an angle θ to the horizontal, are applied to a block of mass m=2.5 kg. The coefficient of kinetic friction between the block and the surface is μk=0.2. The block is moving to the right.

(a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ=30°.

(b) Solve numerically for the magnitude of the acceleration of the block, a in m/s2, if θ=30°.