hree forces acting on an object are given by F, = (-1.9î + 7.45j) N, F, = (4.70î - 1.5j) N, and F, = (-42î) N. The object experiences an acceleration of magnitude 4.00 m/s?. (a) What is the direction of the acceleration?

hree forces acting on an object are given by F, = (-1.9î + 7.45j) N, F, = (4.70î - 1.5j) N, and F, = (-42î) N. The object experiences an acceleration of magnitude 4.00 m/s?. (a) What is the direction of the acceleration?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter4: The Laws Of Motion

Section: Chapter Questions

Problem 8P: A 3.00-kg object is moving in a plane, with its x and y coordinates given by x = 5t2 1 and y = 3t3...

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Concept explainers

Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

Three forces acting on an object are given by F, = (-1.9î + 7.45j) N, F, = (4.70î - 1.5j) N, and
= (-42î) N. The object experiences an acceleration of magnitude 4.00 m/s?.
(a) What is the direction of the acceleration?
• (counterclockwise from the +x-axis)

Expert Solution

Step 1

$\vec{F_{1}}$ =-1.9 i + 7.45 j

$\vec{F_{2}}$ =4.7 i - 1.5 j

$\vec{F_{3}}$ = -42 i

$\vec{F_{T o t a l}}$ = -39.2 i + 5.95 j

| $\vec{F_{T o t a l}}$ | = 39.6489911599 N. $∴$ m = F/a = 39.6489911599/4=9.9122475 Kg

$\vec{a} = \vec{F} / m =$ -3.95470351199 i + 0.6 j

$∴$ direction of acceleration = tan^-1[0.6/-3.95470351199] = -8.62702106907= 171.373^o anticlockwise from x-axis

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