A weight w is to be moved from the bottom to the top of an inclined plane of inclination `theta` to the horizontal. If a smaller force is to be applied to drag it along the plane in comparison to lift it vertically up, the coefficient of friction should be such that.

Options

A. `mugt tan((pi)/4-(theta)/2)`
B. `mult tan((pi)/4-(theta)/2)`
C. `mu lt tan theta`
D. `mugt tan. (pi)/(4)`

Explanation

To solve the problem step by step, we need to analyze the forces acting on the weight \( W \) when it is moved up an inclined plane at an angle \( \theta \) to the horizontal. We will compare the force required to lift the weight vertically with the force required to drag it up the incline.

### Step 1: Identify the force required to lift the weight vertically
When lifting the weight vertically, the force \( F_p \) required is equal to the weight \( W \):
\[ F_p = W \tag{1} \]

### Step 2: Analyze the forces acting on the weight on the inclined plane
When dragging the weight up the inclined plane, we need to consider the components of the weight acting parallel and perpendicular to the incline. The weight \( W \) can be resolved into two components:
- The component acting parallel to the incline: \( W \sin \theta \)
- The component acting perpendicular to the incline: \( W \cos \theta \)

The normal force \( N \) acting on the weight is equal to the perpendicular component of the weight:
\[ N = W \cos \theta \tag{2} \]

### Step 3: Determine the frictional force
The frictional force \( F_v \) opposing the motion up the incline is given by:
\[ F_v = \mu N = \mu W \cos \theta \tag{3} \]
where \( \mu \) is the coefficient of friction.

### Step 4: Set up the equation for the force required to drag the weight up the incline
The total force \( F_d \) required to drag the weight up the incline must overcome both the gravitational component and the frictional force:
\[ F_d = W \sin \theta + F_v = W \sin \theta + \mu W \cos \theta \tag{4} \]

### Step 5: Compare the forces
According to the problem, the force required to drag the weight \( F_d \) must be less than the force required to lift it vertically \( F_p \):
\[ F_p > F_d \implies W > W \sin \theta + \mu W \cos \theta \]

### Step 6: Simplify the inequality
Dividing through by \( W \) (assuming \( W > 0 \)):
\[ 1 > \sin \theta + \mu \cos \theta \]
Rearranging gives:
\[ \mu \cos \theta < 1 - \sin \theta \tag{5} \]

### Step 7: Solve for the coefficient of friction \( \mu \)
Now, we can express \( \mu \):
\[ \mu < \frac{1 - \sin \theta}{\cos \theta} \]
This can be rewritten in terms of tangent:
\[ \mu < \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \tag{6} \]

### Conclusion
Thus, the coefficient of friction \( \mu \) should satisfy the condition:
\[ \mu < \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \]