Question

A block of mass and equals 15.5 kg rests on an inclined plane the coefficient ecstatic friction of .11 between the block and the plane. The inclined plane is L equals 6.8 m long and it has a height of H 3.05 at its tallest point. A)what angle in degrees does the plane make with respect to the horizontal? B. What is the magnitude of the normal force that acts on the block in newtons? C. What is the component of the force of gravity along the plane in newtons?

Answer 1

Step-by-step explanation

Step 1

angle of the inclined plane

to find the angle, we can use the sin function

`\sin\theta=\frac{opposite\text{ side}}{hypotenuse}`

so

a)let

`\begin{gathered} hypotenuse(the\text{ longest side\rparen=6.8} \\ opposite\text{ side\lparen the one in front of the angle\rparen=3.05} \end{gathered}`

b) replace and solve for the angle

`\begin{gathered} \sin\theta=\frac{opposite\text{ side}}{hypotenuse} \\ sin\theta=(3.05)/(6.8) \\ \theta=\sin^(-1)((3.05)/(6.8)) \\ \theta=26.65° \end{gathered}`

therefore

the angle is 26.65 degrees

Step 2

Magnitude of the normal force

a) Diagram

Free body diagram

now ,as the object is at rest, the sum of the forces acting on it must equals zero , hence

for y -axis

`\begin{gathered} Normal\text{ force-mgcos}\theta=0 \\ \end{gathered}`

replace and solve for normal force

`\begin{gathered} Normal\text{ force-mgcos}\theta=0 \\ Normal\text{ force=mg cos}\theta \\ Normal\text{ force=15.5 kg*9.8}(m)/(s^2)cos26.65 \\ Normal\text{ force=131.38 Newtons} \end{gathered}`

so, the Normal force is 131.38 Newtons

Step 3

finally, the component of the force of gravity

to know this, let's set the equation in the x-axis

so

`\begin{gathered} Ff-mg*sin\theta=0 \\ Ff=mgsin\theta \\ (mg\text{ sin}\theta)\rightarrow component\text{ of the force of gravity in the plane } \\ so \\ component=\text{ 15.5kg*9.8}(m)/(s^2)sin26.65 \\ component\text{ =68.13 Newtons} \end{gathered}`

so, the component of the force of gravity along the plane is 68.13 Newtons

I hope this helps you