Question

12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦below thehorizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20m/s2. What is the magnitude of the contact force exerted on the block by the surface?

Answer 1

Answer: Fr = 26.53 N

Explanation: The constant force exerted on the block by the surface is the frictional force.

This frictional force is as a result of interaction between the body and the surface.

According to newton's second law of motion,

F - Fr = ma

F=applied force

Fr = magnitude of frictional force

m = mass of object = 12kg

a = acceleration of object = 3.2m/s²

The applied force (F= 75 N) is inclined at an angle of 30° to the horizontal thus making it have 2 components of forces given below

Fx = 75 * cos 30 = 64.95 N (horizontal component)

Fy = 75 * sin 30 = 37.5 N ( vertical motion)

The body moves across the surface, hence the horizontal component of force is responsible for motion.

F = 64.95 N

By substituting the parameters, we have that

64.96 - Fr = 12 * 3.2

64.96 - Fr = 38.4

Fr = 64.96 - 38.4

Fr = 26.53 N

Answer 2

Answer:

26.55N

Explanation:

Newton's second law of motion states that the effective/resultant (∑F) force acting on a body is the product of the mass (m) of the body and the acceleration (a) caused by this force, as follows;

∑F = m x a

From the question, the two forces acting on the body are;

(i) the force (say F) of push which is 75.0N inclined at 30° below the horizontal. Since this force is inclined at an angle, it means that it has two components - the vertical ($F_{Y}$) and the horizontal ($F_{X}$) - and are calculated as follows;

=> $F_{Y}$ =  F sin θ

Note: That the vertical component, $F_{Y}$, of the force will be negative since it is inclined at an angle below the horizontal

∴ $F_{Y}$ =  -75.0 sin 30° =  - 37.5N

Also,

=> $F_{X}$ =  F cos θ

∴  $F_{X}$ =  75.0 cos 30° = 64.95N

Since the motion of the block is on the horizontal, the horizontal component ($F_{X}$ = 64.95N) is sufficient to solve the problem at hand and will be used.

(ii) the contact force exerted on the block by the surface. This force is due to friction and is called the frictional force ($F_{R}$) and it acts to oppose the force of push on the block. Therefore, it is going to have a negative sign to show that it acts in a direction opposite to that of the force of push.

Therefore, the resultant/effective force (∑F) is found by summing the horizontal component of the force of push ($F_{X}$) and the frictional force ($F_{R}$) which also acts in the horizontal, as follows;

∑F = $F_{X}$ - $F_{R}$

∑F = 64.95 - $F_{R}$

Now, substitute the value of ∑F into equation (i) as follows;

64.95 - $F_{R}$ = m x a              ----------------------(ii)

Where;

m = mass of the block = 12.0kg

a = acceleration of the block = 3.20m/s²

Substitute these values into equation (ii) as follows;

64.95 - $F_{R}$ = 12.0 x 3.20

64.95 - $F_{R}$ = 38.4

Solve for $F_{R}$;

$F_{R}$ =  64.95 - 38.4

$F_{R}$ = 26.55N

Therefore, the contact force exerted on the block by the surface is 26.55N