Sign up for our free STEM online summer camps starting June 1st!View Summer Courses

Problem 33

$\mathrm{A} 1000 \mathrm{kg}$ boat is traveling a…

02:20

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 32

A block is pushed across a floor by a constant force that is applied at downward angle $\theta($ Fig. $6-19) .$ Figure $6-36$ gives the acceleration magnitude $a$ versus a range of values for the coefficient of
kinetic friction $\mu_{k}$ between block and floor: $a_{1}=3.0 \mathrm{m} / \mathrm{s}^{2}, \mu_{k 2}=$
$0.20,$ and $\mu_{k 3}=0.40 .$ What is the value of $\theta ?$

Answer

$\theta=60^{0}$


Topics


Discussion

You must be signed in to discuss.

Video Transcript

So let's draw first the free body diagram for the system. We have a point. Mass will treat it as a point ness. Uh, going up is forced. Normal. Going down is you wait mg at an angle. We have f making an angle with the horizontal Seita and opposite to the direction of motion would be the force of fiction. At this point, we can apply the sum of forces in the ex direction. This will equal the mass times acceleration. This would equal f co sign of data minus the force of friction. We can say that some of forces in the UAE direction would be equal to forced normal minus the force sign if ADA minus M. G, this is equaling zero because there isn't any acceleration in the UAE direction we could we know that here the force of friction would be equal to the coefficient of kinetic three times the force normal and according to this equation, this will be equal to the cowfish for of the coefficient of friction, the kinetic coefficient of friction multiplied by M G plus F sign off data. So at this point, we can then substitute this into the first equation and say f co sign of Fada minus the coefficient. Kinetic friction times M G minus f Sign of data would be equal to the mass times acceleration and so we can solve for the acceleration. This is equaling F over m co sign of Fada minus the coefficient of kinetic friction times Sign of data. This is equaling rather minus the coefficient of friction times g from the figure we know that a is equaling 3.0 ah, 3.0 meters per second squared and we know that the coefficient of kinetic friction is equal to zero. So this essentially means that 3.0 meters per second squared, it would be equal to f co sign of data divided by M Um, we also know that at a equaling zero meters per second and the coefficient of kinetic friction equaling 0.20 we can say that zero equals f over and Times co sign of theta minus 0.2. Sign of fada. Close parentheses minus, Uh, point. Okay, let's through type. That's given your workbook from these parameters. Here we can say that zero equals f over M Times Co sign of data minus 0.20 Sign of Fada closed parentheses minus 0.20 times 9.8 meters per second squared. And so at this point, we can then say that this F over m co sign a fada minus 0.20 sign of Fada close parentheses minus 0.20 times 9.80 meters per second squared is gonna be equal to 3.0 meters per second squared minus 0.20 I've been saying minus 0.20 f over M sign of fada minus 1.96 meters per second squared. We condone say that zero is gonna be equal to 1.4 meters per second squared minus 0.20 times f sign of Fada divided by M and we can then say 5.2 zero meters per second squared equals f sign of Fada over m and we're going to combine the two results combining this with this equation. We find that tangent of Fada will equal 5.2 meters per second squared, divided by 3.0 meters per second squared. This is equaling 1.73 fatal with n equal our time of 1.73 This is giving us approximately 60 degrees. So this is our angle theta. That would be our final answer. That is the end of the solution. Thank you for watching.

Recommended Questions