# Physics - Newton's 3rd Law Question?

Blocks X and Y are attached to each other by a light rope and can slide along a horizontal, frictionless surface. Block X has a mass of 10 kg and block Y a mass of 5.0 kg. An applied force of 26 N [right] acts on block X.

(a) Calculate the action-reaction forces the blocks exert on each other.

(b) Suppose the magnitudes of the force of friction on blocks X and Y are 8.0 N and 4.0 N respectively. Calculate the action-reaction forces the blocks exert on each other.

I understand how to solve problems with the first two laws, but the problems that involve the 3rd law are tricky for me. I know that the equation used is: F = ma and that F(x on y) = - F (y on x) - for example. I'm just not sure how to go about this question.

I'd appreciate your help on helping me solve this problem.

### 2 Answers

- oldprofLv 78 years agoFavorite Answer
OK we go [Y]===[X]====> F; where m = 5 kg and M = 10 kg. F = 26 N.

The sum of forces is zero according to Newt; so F - (m + M)A = 0 = F - Fy - Fx = 0 where Fy = mA and Fx = MA are the two reaction forces on m and M respectively. And of course the reaction in this case is for the two masses to accelerate A in response to yanking on the rope. The acceleration here is A = F/(m + M) = 26/(5 + 10) = 26/15 m/s^2. ANS. a.

We add friction; so the sum is now F - fy - fx = (m + M)A [I put the acceleration term on the RHS for convenience, but the sum is still zero when everything in on the LHS.] Now we have 26 - 12 = (m + M)A and A = 14/15 m/s^2 is the acceleration. ANS. b.

- begnocheLv 44 years ago
1) the web drive of any process is always equal to mass times acceleration. Therefore, the magnitude of the drive is simply 2 occasions 6 which is 12N. For the 2d section, the equal drive is applied so 12N = 4kg(a). Acceleration in this case would be three m/s^2 2. There can be acceleration for the reason that the only force that we're concerned with on this case is the drive of the baby on the wagon. If we take the approach to be best the wagon, we are able to comfortably see how there will probably be acceleration. We are able to additionally keep in mind the example of an object shedding. You wouldn't don't forget the earth quite often would you? 3a) To take a step, one have to push backwards and its friction that pushes one forwards. B) A snowball hitting any individual within the back, the again pushing the snowball. C) A baseball participant catches a ball, the ball pushes the baseball player. D)A gust of wind strikes a window, the window resists the gust. Four. First calculate the magnitude of acceleration. To do that, we'd first ought to do some vector addition. Keep in mind that force x squared plus drive y squared = resultant force squared. The ensuing drive on this case could be 430N. To find acceleration, we say force net = ma. Web force on this case is 430N so 430N = 270kg(a). A = 1.59 m/s^2. The direction can also be determined by taking the arc Tangent. The course is sixty five.2 degrees north of east.