gn Plot 0.8 L 64° ....*.

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As shown below, on Venus a 29 kg beam of length L is leaning against a frictionless wall and is in static equilibrium. The center of mass of the beam is located 0.8L from the lower end of the beam. Determine the force the wall exerts on the beam and the minimum coefficient of static friction between the floor and the beam.

The image shows a diagram of a rectangular block leaning against a vertical wall, forming a right triangle with the floor. The block is outlined in orange and includes a red dot on its surface, indicating a point of interest. The length of the block is labeled as \( L \).

Key elements of the diagram:
- The block is inclined at an angle of \( 64^\circ \) with the horizontal floor.
- A dashed line is drawn from the base of the wall to a point on the block, labeled as \( 0.8L \), indicating the horizontal distance from the wall to the block along the dashed line.
- The angle between the block and the horizontal is marked in blue.

This setup is typically used in physics problems involving forces, inclines, or stability analysis.
Transcribed Image Text:The image shows a diagram of a rectangular block leaning against a vertical wall, forming a right triangle with the floor. The block is outlined in orange and includes a red dot on its surface, indicating a point of interest. The length of the block is labeled as \( L \). Key elements of the diagram: - The block is inclined at an angle of \( 64^\circ \) with the horizontal floor. - A dashed line is drawn from the base of the wall to a point on the block, labeled as \( 0.8L \), indicating the horizontal distance from the wall to the block along the dashed line. - The angle between the block and the horizontal is marked in blue. This setup is typically used in physics problems involving forces, inclines, or stability analysis.
Expert Solution
Step 1

Given,

Mass, m=29kg

Weight, 

W=mgW=29×8.87W=257.23N

Advanced Physics homework question answer, step 1, image 1

Let the force exerts on the wall be F.

Fx=0F=fF=μsNFy=0W=NN=257.23N

W×0.8Lcos64°=F×Lsin64°257.23×0.8cos64°=F×sin64°F=100.36N

 

 

 

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