Chapter 5, Problem 1Q

Figure 5-19 gives the free-body diagram for four situations in which an object is pulled by several forces across a frictionless floor, as seen from overhead. In which situations does the acceleration $\stackrel{\to }{a}$ of the object have (a) an x component and (b) a y component? (c) In each situation, give the direction of $\stackrel{\to }{a}$ by naming either a quadrant or a direction along an axis. (Don’t reach for the calculator because this can be answered with a few mental calculations.)

To determine

To Find

a) Which situation have x component of acceleration.

b) Which situation have y component of acceleration.

c) Direction of acceleration for each situation.

Answer to Problem 1Q

Solution

a) 2, 3 and 4.

b) 1, 3 and 4.

c) 1 – Along + y-axis, 2- Along + x-axis, 3- In 4^th quadrant and 4- In 3^rd quadrant.

Explanation of Solution

1) Concept:

Using the concept of net force from the Newton’s second law of motion, we can find the net force acting on the given object for given conditions.

2) Calculations:

a) According to Newton’s second law net force is product of mass and acceleration.

If we want x component acceleration there must be net force in x direction

So, For situation 1

Net force in x direction

$F_{net}=5-3-2=0$.

So, there is no x component of acceleration.

For Situation 2

Net Force in x direction

$F_{net}=3-2=1 N$.

As net force is 1N, x component of acceleration is present.

For Situation 3

Net Force in x direction

$F_{net}=5-4=1N$.

As net force is 1N, x component of acceleration is present.

For Situation 4

Net Force in x direction

$F_{net}=3-5=-2 N$.

As net force is 1N, x component of acceleration is present.

b)

For situation 1

Net force in y direction

$F_{net}=7-4=3$.

So, there is y component of acceleration.

For Situation 2

Net Force in y direction

$F_{net}=6-2-4=0 N$.

As net force is no y component of acceleration is present.

For Situation 3

Net Force in y direction

$F_{net}=6-7=-1N$.

As net force is -1N, y component of acceleration is present.

For Situation 4

Net Force in y direction

$F_{net}=3+2-5-4=-4 N$.

As net force is -4N, y component of acceleration is present.

c) Direction of acceleration is in direction of net force.

For situation 1 there is only net force is only in +y direction so acceleration is also in +y direction.

For situation 2 there is only net force is only +x direction so acceleration is also +x direction.

For situation 3 as there is net force both in x and y direction and total net force is in fourth quadrant.

 For situation 4 as there is net force both in x and y direction and total net force is in third quadrant.

Conclusion:  Using the equations from the Newton’s second law of motion and vector algebra, it is possible to find the net force acting on the system.