A block of weight W rests as shown in a wedge of negligible weight

W=106.825lb

N1=118.189lb

Ff1=29.547lb

N2=122.846lb

Ff2=30.712lb

N3=45.453lb

Ff3=11.363lb

Kindly see the photo below and explanation. Thank you very much.

Step-by-step explanation

Definition:

Impending motion is a motion where the object starts to slide.

Frictional force is a resisting force that is opposite to the force applied to an object.

Angle of friction is an angle between the normal force and the frictional force.

Given:

Coefficient of static friction= 0.25

P=75 lb

Solution:

We first draw the free body diagram of the second block because the given force is on the second block. The force 75lb moves the second block to the left so the frictional force will be to the right direction because it is resisting the movement of the block. Normal force is on upward direction and the frictional angle will be inverse tangent 0.25° = 14.036°. There will be resultant between the normal and frictional force and this force will be relative on the triangle force method.

After drawing the free body diagram and its angles, we now proceed to the triangle force method. The direciton of the forces P, R1 and R2 present in the second block will result in a triangular form if joined together following the tails of the forces.

Using sine law method, we will know the values for R1 and R2.

R1=75sin(68.964) /sin(35.072) = 121.826lb

R2=75sin(75.964)/sin(35.072)=126.627lb

On the first block, we will draw the free body diagram including its relative forces such as N2, Ff2, R2, W, N3, Ff3 and R3 along with its angles.

We will now draw the triangle force method. The forces R2, R3 and W will result in a trinagular force following the force's tails. We will use sine law in finding the values for W and R3.

W= 126.627sin(54.928°)/sin(104.036°)=106.825lb

R3=126.627sin(21.036°)/sin(104.036°)=46.852lb

After computing the resultant 1, 2 and 3, we can now know the values for the normal forces and frictional forces by using trigonometric functions.

N1=121.826cos(14.036°)=118.189lb

N2=126.627cos(14.036°)= 122.846lb

N3=46.852cos(14.036°)= 45.453lb

For finding the frictional forces we wilk substitute the values for normal forces on the formula for frictional force= uN

Ff1=0.25×118.189=29.547lb

Ff2=0.25×122.846=30.712lb

Ff3=0.25×45.453=11.363lb

Where: Ff= frictional force

N= normal force

u= coefficient of static force

Theta= angle of friction

R = resultant of normal force and frictional force

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