A rock is on a slope at an angle of 31.4 degrees, with respect to the horizontal

Question

A rock is on a slope at an angle of 31.4 degrees, with respect to the horizontal. The mass of the rock is 31.8 kg. Remember to draw a picture of the situation, put in all the forces acting on the rock. Then write down Newton's 2nd law equation for the forces in the x direction and in the y direction. (1) What is the normal force of the slope on the rock? (2) Assuming there is no motion what is the friction on the rock? (3) What is the maximum static friction if the coefficient of static friction is 0.330? (4) What is the kinetic friction if the coefficient of kinetic friction is 0.170? (5) What is the acceleration of the rock down the slope?

Explanation / Answer

1) The forces acting on the rock parallel to the slope (the "x-axis", with the coordinate plane slanted at 31.4 degrees), are the component of the force of gravity down the slope (mgsin?) and the force of friction up the slope (-µn). These forces sum to zero because the rock is stationary. From newton's 2nd law: ?F(x) = 0 = mgsin? - µn-------------->(1) Perpendicular to the slope the forces are the perpendicular component of gravity (-mgcos?) and the upward normal force, n. Again, the force sum to zero, and the equation is: ?F(y) = 0 = n - mgcos? n = mgcos?----------------------->(2) So the normal force of the slope on the rock is: n = (31.8kg)(9.80m/s²)cos31.4° = 266 N--> answer 2_ By expressing the friction force (µn) in (1) as f, we have: 0 = mgsin? - f f = mgsin? = (31.8kg)(9.80m/s²)sin31.4° = 162.36N--> answer 3) The maximum force of static friction is found from: f = µn = µmgcos? = 0.330(31.8kg)(9.80m/s²)cos31.4° = 87.8N--> answer 4) The kinetic friction force is found the same way: f = µmgcos? = 0.170(31.8kg)(9.80m/s²)cos31.4° = 45.22 N --> answer 5) If the rock accelerates down the slope, then the net forces are unbalanced and the equation from NSL for the parallel forces is: ?F(x) = ma = mgsin? - µn We know the normal force from (2), and it still holds in the case of the rock rolling down the slope because there is NO movement perpendicular to the slope. So the last equation becomes: ma = mgsin? - µmgcos? Solved for acceleration: a = g(sin? - µcos?) = 9.80m/s²(sin31.4° - 0.170cos31.4°) a= 3.7m/s²--> answer

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