Another homework question for my kid. Please help!!!!!!!!!!?

Question

A 70 kg box is pulled across a frictionless surface by a 300 N force at an angle of 30 degrees to the horizontal.

a. What is the acceleration of the box in the x-direction?

b. What is the normal force on the box?


Please show work.

Answer 1

Normal force is perpendicular to the 30 deg. plane. If the box was on the ground the normal force would be the mass, but it is at a 30 deg angle so you must multiply the mass by sin(30) to obtain your new normal force.

You can use one of Newton's laws F=ma. Force= Mass* acceleration. So therefore Force/mass = acceleration on the 30 deg. plane. In order to get the x-direction you will need to take that number and multiply it by cos (30). I am not going to do the calculations for you, but that will be the answer.


The trick is knowing triginometry and what a right triangle is.

Answer 2

a. well since F=M*A ==> force equals mass * acceleration
then
300 N * cos(30(degrees))=70 kg * A

notice that 300 N is multiplied by cos(30) to account for inclination

Hence A= 300 N * cos(30) / 70 kg

* Convert neutons to kg and there is your answer

Normal force is the force applied up perpendicular to inclination
thus, 300 N * sin(30(degrees))

** double check in book im kinda rusty i could have mistaken sin for cos, but essentially this is how you solve this problem :)