What Force Keeps the Block Stationary on a Frictionless Wedge?

[southernguy13]

Homework Statement

The surfaces between a 11 kg block, the 16 kg wedge and between the 16 kg wedge and the horizontal plane are smooth (without friction). The acceleration of gravity is 9.8 m/s2 .
A block is released on the inclined plane (top side of the wedge). What is the force F which must be exerted on the 16 kg block in order that the 11 kg block does not move up or down the plane? Answer in units of N.

Homework Equations

F=ma

The Attempt at a Solution

I drew a free body diagram with the i axis horizontal and the j axis vertical for each object, the block and the ramp. The block has a force of gravity, in the negative j direction and a normal force in the positive j direction. The acceleration of the block is 0.

I'm not sure how to figure out the exact free body for the ramp, since it's angled, or how to find what the force should be.

Thanks for the help!

 

[rl.bhat]

When you apply a force F on the larger block, it will accelerate a towards left. It will cause an a acceleration on the smaller block towards right. Resolve this acceleration into two components. One along the inclined plane in the upward direction and another perpendicular to the inclined pale. Similarly weight of the smaller block is resolved into two components. One along the inclined plane in the downward direction. and another perpendicular to the inclined plane. Now what should be the condition that the smaller block will remain at rest?

 

[tomcue]

I would approach this problem by using the equation F=ma, where F represents the net force on the object, m is the mass of the object, and a is the acceleration of the object. In this case, the object is the 11 kg block and the acceleration is 0. Therefore, the net force on the block must also be 0 in order for it to stay in place on the ramp.

To find the force exerted on the 16 kg wedge, we can use the fact that the normal force on the block is equal in magnitude to the force exerted on the wedge by the block. This is because the two objects are in contact with each other and the normal force acts as a reaction force to the force exerted by the block.

Using this information, we can set up an equation: F - N = ma, where F is the force exerted on the wedge, N is the normal force, m is the mass of the wedge, and a is the acceleration of the wedge. Since the wedge is not accelerating (it is in equilibrium), a is also equal to 0. Therefore, the equation simplifies to F = N.

To find the normal force, we can use trigonometry to break down the weight of the block into its components along the x and y axes. The weight of the block is mg, where m is the mass of the block and g is the acceleration of gravity. The component of the weight along the ramp is mg sinθ, where θ is the angle of the ramp. Therefore, the normal force is equal in magnitude to mg sinθ.

Putting it all together, we can write the equation as F = mg sinθ. Plugging in the given values, we get F = (11 kg)(9.8 m/s^2)(sin 30°) = 53.9 N.

Therefore, the force that must be exerted on the 16 kg wedge in order for the 11 kg block to stay in place is 53.9 N.