Keep the dog safe

Given, normal force=mg=98N

So, mass= $\frac{98}{g}$ =10kg

Now, acceleration is 2 m/ $s^{2}$

So force of friction is $m*a$ = $10*2$ = $\boxed{20N}$

The weight of the dog is 98N = mass of dog $\times$ 9.8 So the mass of the dog is 10 Kg. When the truck moves forward, due to inertia the dog should move backwards with the same acceleration $2 m/s^{2}$ .

The force can be calculated by F= m $\times$ a

F = 10 $\times$ 2

F = 20N

The frictional force (f) must be acting opposite to this force

We know that the dog remains stationary so the net force must be = 0

$F_{net}$ = 0

F - f = 0

f = F

Thus the frictional force, $\boxed{f=20N}$

We know that $\sum F=ma$ .

The dog is stationary, so the Force of Friction must be equivilant to the Force on the dog from the acceleration of the truck.

$F_{f}=ma$

Mass is not given, so we must calculate mass.

$F_{g}=mg$

$m=\frac{F_{g}}{g}$

$m=\frac{98 N}{9.8 m/s^{2}}$

$m=10kg$

The acceleration is given as $2 m/s^{2}$

$F_{f}=(10 kg)(2m/s^{2})$

$F_{f}=20 N$

$\boxed{20N}$

As the dog is not moving even after the truck accelerates so there is no use of kinectic friction...i cant even find any use of static friction...its simple that the dog has 98/9.8 = 10 kg mass.And when the truck starts to accelerate it resists 2*10 = 20N force and stay still.So the answer is 20N

Though I am writing a solution, I am confused a little bit: First solution: :)

In dog's frame, since dog is stationary, centrifugal force is balanced by force of friction. Therefore, force of friction must be $m*a$ where $m$ is the mass of dog. This mass is given by $\frac{W}{g}$ where $W$ is weight of the dog. Thus $m=\frac{98}{9.8}=10$ and hence required force is $10*2=20 N$ .

Now about my confusion. The force of static friction on the dog should not depend on any other force acting on dog in my view. So independent of centrifugal force, it should be equal to $\mu N=0.8*98=78 N$ . What is wrong with this?

The only horizontal force on the dog is the force of friction. The horizontal acceleration of the dog is $2 \ \mathrm{m} / \mathrm{s}^2$ , and its mass is equal to the quotient of its weight and $g$ . The mass is therefore $10 \ \mathrm{kg}$ . Thus, the frictional force is the mass times the acceleration, or $\boxed{20 \ \mathrm{N}}$ .

first calculate the total static frictional force = coefficient of static friction * normal reaction which turns out to be 78.4. now calculate the force due to the moving of the truck on dog . mass of the dog = normal force / accelaration due to gravity = 10kg. .force on the dog = mass * accelaration = 20 Newtons . only 20 Newtons is sufficient to prevent the dog from falling off the truck . so,the static frictional force adjusts itself to 20 Newtons .

Mass of dog is (98/g)=10 kg

Car moves at 2 m/s so dog moves at 2 m/s.

Force on dog is ma=10*2==20

Cute Puppy

98=mg,so m=10. right side ma=10.2=20.the dog is not moving. so force of friction is also 20 newton

the dog won't move if the force of static friction has the same or higher value than the force given by the truck. So: M=N/g=98/9.8=10 kg; F=Ma= 10*2=20N :))

force of static friction (fs) = μR, where R is the normal reaction force which is equal to 98 N.

                                              =0.8 * 98 N

R=mg, where m= mass of the body.

fs= μR

fs= μmg

0.8 * 98 =0.8 * m * 9.8

m= (0.8 * 98)/(0.8 * 9.8)

= 10 kg

F=ma, where a=2 m/s(given).

=10 * 2

= 20 N

The coefficient of friction being stated is irrelevant and a red herring- we could only use it if we knew that the that friction is acting during limiting equilibrium.

the mass of the dog is $98/9.8=10$

Since there is no sliding friction is providing all the force required to accelerate the dog with the car hence simply using

$F=ma$ we have $F_r=2*10=\boxed{20}$

In fact using the coefficient of friction will give $F_r=\mu R= 0.8*98=78.4$ so you are way under $LE$

2 (98/9,8) = 2 10 = 20 N

Because the dog does not slide when the truck accelerates forward, the dog must also accelerates at the same rate as the truck, which is $2 ms^{-2}$ . This acceleration is caused by the friction between the dog and the truck, which can be calculated using the formula:

$F = m \times a$

with $m = \frac{W}{g} = \frac{98}{9.8} = 10 (kg)$

Therefore:

$F = 10 \times 2 = 20 (N)$

The coefficient of static friction between the dog and the truck is not needed.

the dog is not slide so f=ma a=gs so we can calculate this f=98/9.8*2=20

1. mass of dog = weight in newtons/acceleration due to gravity = 98/9.8 = 10 kg
2. force doggie experiences when truck accelerates = mass x acceleration = 10 x 2 = 20 N
3. now simply put, since Mr. Doggie stays put in his place, it is safe to say that frictional force is equal to the force experienced during acceleration ,since an equal force acts in a direction opposite to that of motion ( that is, towards right & left respectively)

La aceleracion del perro es 2 m/ s¨2, la masa del perro es 10 kg, ya que el peso es 98N y la gravedad es -9.8 m/s2 y m = peso / gravedad , entonces la fuerza resultante sera masa x aceleracion = 10 m/s2 x 2 kg = 20 N que tambien es la fuerza de rozamiento

Weight of dog= $m\times g$ =98N, m= $\frac{98}{9.8}$ =10kg. Since the dog is stationary, the net force on it is zero. A frictional force is acting in a direction opposite to the truck's motion on the dog to prevent it from sliding. F=ma= $10\times2$ = $\boxed{20N}$

F=ma. The net force acting on the dog is friction from the car. Hence, just solve by using m = W/g and the a is given

It says it's on the truck and doesn't move even accelerating. So, just get the mass of the dog then times with the acceleration because the force exerted is same as the frictional force. Exerted F = Friction F, so... Friction = (98/9.81)x 2... F= 19.98

Let mass be m, friction force be f.

mg = 98

Therefore, m = 10kg.

Pseudo force acting on dog = 10 * 2 = 20N

f(max) = 10 * 9.8 * 0.8 = 78.4N that can act on dog.

But as applied pseudo force is less than f(max), then force of friction acting on dog = 20N , i.e., pseudo force. (Ans)