A 5.7 kg bag of groceries is in equilibrium on an incline of angle = 18°. Find the magnitude of the normal force on the bag.
____ N
i found the components already and i don't know what to do next.
2006-10-18 13:05:43 · 5 answers · asked by tingerpoo 2 in Science & Mathematics ➔ Physics
I don't have a calculator with me. Normal force is always perpendicular to the surface the object is on. The normal force is going to counteract what force is opposite to it. which is one of the components of mg. Which would turn out to be (cos18)(m)(g) = normal force.
therefore normal force = (cos18)(5.7)(9.8)
normally it would be -cos18mg becaue it is opposite but i just considered g to be positive instead since it would cancel out.
2006-10-18 13:11:03 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Hi. Just calculate the normal force (down on the surface) in Newtons.
2006-10-18 13:10:00 · answer #2 · answered by Cirric 7 · 0⤊ 0⤋
Normal Force = mgcos(theta)
= (5.7 kg) (9.89m/s^2) cos (18)
2006-10-18 13:17:09 · answer #3 · answered by Phillip 3 · 0⤊ 0⤋
ok. your next step is that since it is in equilibrium the normal force is equal to the y component of the force of gravity. so, Fgravity in y = normal force
2006-10-18 13:08:58 · answer #4 · answered by life_television 1 · 0⤊ 0⤋
If i still can recall is 5.7 x 9.81 x cos 18°
2006-10-18 13:25:43 · answer #5 · answered by Mr. Logic 3 · 0⤊ 0⤋