Physics - Calculating Force

Franklin lifts a weight vertically. If he lifts it so that way the velocity of the weight is constant, how much force is he using?

$\small g=-9.8\frac{m}{s^2}$

Explanation

If the object has a constant velocity, that means that the net acceleration must be zero.

$a=\frac{v_2-v_1}{t}$

$v_1=v_2\rightarrow a=\frac{0}{t}=0$

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

$F_{net}=ma_{net}=m(0\frac{m}{s^2})=0N$

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

$F_{net}=F_{lift}+F_{gravity}$

Since we just proved that the net force will equal zero, we can say $-F_{gravity}=F_{lift}$ .

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

$-F_{gravity}=F_{lift}$

$-(mg)=F_{lift}$

$-(-9.8\frac{m}{s^2}*2kg)=F_{lift}$

$-(-19.6N)=F_{lift}$

$19.6N=F_{lift}$

A dog bites down on a chew toy with of force. What is the force of the chew toy on the dog?

We need more information to solve

Explanation

Use Newton's third law to solve this question. The force of the dog on the chew toy is equal in magnitude, but opposite of direction, to the force of the chew toy acting on the dog.

That means that $F_{dog}=-F_{chew toy}$ .

Using our given values for find the force of the chew toy.

$3N=-F_{chewtoy}$

$-3N=F_{chewtoy}$

Screen_shot_2013-12-27_at_1.15.43_am

If the mass of the object is and $\theta=30^\circ$ , what is the normal force on the object?

$\small g=-9.8\frac{m}{s^2}$

Explanation

The normal force is always perpendicular to the surface upon which the object is moving, and is pointed away from said surface. That means we are looking for the value for Z in the diagram.

Observe that Z and Y are equal, but opposite forces.

If we can solve for Y, then we can find Z.

We can use our understanding of trigonometry to find an equation for Y.

$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

If we plug in for the angle, we see:

$\cos(a)=\frac{Y}{W}$

Since we are solving for Y, we can multiply both sides by W.

Now that we know an equation for Y, we can return to our original equation to solve for Z.

$Z=-(W*\cos(a))$

From here, we can use Newton's second law to find the value of W, the total force of gravity.

Substitute this into our equation for Z.

$Z=-(mg*\cos(a))$

Now we can solve for Z using the values given in the question for the angle, mass, and gravity.

$Z=-(30kg*-9.8\frac{m}{s^2}*\cos(30^\circ))$

Jerry pushes a box with of force. What is the resultant acceleration?

$0.47\frac{m}{s^2}$

$0.66\frac{m}{s^2}$

$1.2\frac{m}{s^2}$

$4.78\frac{m}{s^2}$

$0.06\frac{m}{s^2}$

Explanation

Newton's second law states that $\vec{F}=m\vec{a}$ .

In this problem, $\vec{F}=1.5N$ and .

Plug these into the equation to solve for acceleration.

$\vec{F}=m\vec{a}$

$\frac{1.5N}{3.2kg}=a$

$0.47\frac{m}{s^2}=a$

Screen_shot_2013-12-27_at_1.15.43_am

If the mass of the object is and $\theta=41^\circ$ , what is the normal force on the object?

$\small g=-9.8\frac{m}{s^2}$

Explanation

The normal force is always perpendicular to the surface upon which the object is moving, and is pointed away from said surface. That means we are looking for the value for Z in the diagram.

Observe that Z and Y are equal, but opposite forces.

If we can solve for Y, then we can find Z.

We can use our understanding of trigonometry to find an equation for Y.

$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

If we plug in for the angle, we see:

$\cos(a)=\frac{Y}{W}$

Since we are solving for Y, we can multiply both sides by W.

Now that we know an equation for Y, we can return to our original equation to solve for Z.

$Z=-(W*\cos(a))$

From here, we can use Newton's second law to find the value of W, the total force of gravity.

Substitute this into our equation for Z.

$Z=-(mg*\cos(a))$

Now we can solve for Z using the values given in the question for the angle, mass, and gravity.

$Z=-(14.2kg*-9.8\frac{m}{s^2}*\cos(41^\circ))$

Michael pushes a box with of force to the left. Annie pushes the same box with of force to the right. What is the net force on the box?

Explanation

For the net force, we add up all the forces: $F_{net}=F_{Michael}+F_{Annie}$ .

Since force is a vector, the direction of the action matters. We will make leftward motion negative and rightward motion positive. Michael is pushing with to the left, making his force equal to . Annie was pushing with to the right, so her force will remain .

We can find the net force by adding the individual force together.

$F_{net}=F_{Michael}+F_{Annie}$

$F_{net}=-30N+40N$

$F_{net}=10N$

Derek pushes a crate along a rough surface with of force. He observes that it is only accelerating at a rate of $2\frac{m}{s^2}$ . What must the value of the force of friction be?

(Assume the only two forces acting on the object are friction and Derek).

Explanation

Newton's second law states that $\vec{F}=m\vec{a}$ .

If Derek is pushing with of force, then we should be able to solve for the acceleration of the crate.

$35N=(10kg)(\vec{a})$

$\frac{35N}{10kg}=\vec{a}$

$3.5\frac{m}{s^2}=\vec{a}$

Derek observes that the crate is acceleration at a rate of $2\frac{m}{s^2}$ , rather than the expected $3.5\frac{m}{s^2}$ . An outside force is acting upon it to slow the acceleration.

The equation for the net force on the object is: $\vec{F}_{resultant}=\vec{F}_{Derek}+\vec{F}_{friction}$ . We also know, from Newton's second law, that $\vec{F}_{resultant}=m\vec{a}_{resultant}$ , where the resultant force and acceleration are the values actually observed.

Plug in the information we've been given so far to find the force of friction.

$\vec{F}_{resultant}=\vec{F}_{Derek}+\vec{F}_{friction}$

$m\vec{a}_{resultant}=(35N)+\vec{F}_{friction}$

$(10kg)(2\frac{m}{s^2})=(35N)+\vec{F}_{friction}$

$20N=35N+\vec{F}_{friction}$

Subtract from both sides to find the force of friction.

$-15N=\vec{F}_{friction}$

Friction will be negative because it acts in the direction opposite to the force of Derek.

A hammer exerts a force on a nail. What is the force the nail exerts on the hammer?

Explanation

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

Since the hammer exerts of force on the nail, the nail must exert of force on the hammer.