A 2.00 kg block is held in equilibrium on an incline of angle = 70° by a horizon

Question

A 2.00 kg block is held in equilibrium on an incline of angle = 70° by a horizontal force applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is s = 0.300, determine the following.

(a) the minimum value of vector F
_____ N

(b) the normal force exerted by the incline on the block
_____ N

A 2.00 kg block is held in equilibrium on an incline of angle theta = 70 degree by a horizontal force applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is Mus = 0.300, determine the following. (a) the minimum value of vector F _____ N (b) the normal force exerted by the incline on the block _____ N

Explanation / Answer

Start by adding all of the forces along the incline.

Fcos-mgsin+Fs=0 (since it is not accelerating)

Now you need to find Fs which is Fs=s*FN

FN is the force of the incline on the object, so add all of the forces perpendicular to the incline.

FN-mgcos-Fsin=0

FN=mgcos+Fsin

So, Fs=s(mgcos+Fsin)

Now plug this into the original equation and you get,

Fcos-mgsin+s(mgcos+Fsin)=0

Fcos-mgsin+smgcos+sFsin=0

Since you are solving for F, keep them both on the left side.

Fcos+sFsin=mgsin-smgcos

F(cos+ssin)=mgsin-smgcos

F=(mgsin-smgcos)/(cos+ssin)

Plug number in and you get,

a.) F=26.3 N

FN was already found earlier to be

FN=mgcos+Fsin

Plugging the answer to a (F=26.3 N) into this equation gives,

b.) FN=31.4 N

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