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A box with a mass of 10 $\mathrm{kg}$ is placed on an inclined plane that makes a $60^{\circ}$ angle with the horizontal. The coefficient of static fraction $\left(\mu_{\mathrm{s}}\right)$ between the box and the inclined plane is $0.2 .$ What force must be (A) 32.7 $\mathrm{N}$(B) 48.3 $\mathrm{N}$(C) 56.2 $\mathrm{N}$(D) 76.6 $\mathrm{N}$applied on the box in order to prevent the box from sliding down theinclined plane?

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Chapter 4

Practice Test 4

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their forces on a box on the inclined plane are following. So we have a box here, the force downward. He's if wait, scientific data scientist that where the force of friction is outward. So force of friction here and the normal force in this direction and downward. He's the the compliment off the wait. So where it is here? The compliment of right here is if the beautiful wait cost that I see then equating the forces on X axis and why direction? Access. Let's say this use of our X and this is why So why? Then we can write our forces. Someone forced on the box is equal to zero. We can write. It's f normal is a call to if normal is equal to force. Wait times course later course, Vicar. Um then we can write a forces on why Direction? Which our effort Clyde will be cool too. N g scientific minus the forces of friction signs minus a mule mg n g cost data course later, that's obstructing the values. Um, we have a mass is a 10 traditional exertions. A 10 sign off 60 sign off 60 year and minus use little 0.2 when deployed by the Mass will be over 10 times 10. I didn't times course or for 60 of course. Over. A six see 60. Then they applied force. We get by simplifying this expression is 76.6 Newton. Then, looking into our options, we see Option D is the right answer is the right answer.

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