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A cart weighing 40 lb is placed on a ramp inclined at 15° to the horizontal. The cart is held in place by a rope inclined at 60° to the horizontal, as shown in the figure. Find the force that the rope must exert on the cart to keep it from rolling down the ramp.

A cart weighing 40 lb is placed on a ramp inclined at 15° to the horizontal. The cart-example-1
User DavidRguez
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1 Answer

1 vote

Answer:

11.97 lb

Explanation:

To find the force that the rope must exert on the cart to keep it from rolling down the ramp, we need to resolve the forces acting on the cart along the direction of the ramp and perpendicular to the ramp.

First, we resolve the weight of the cart into its components. The weight of the cart acting vertically downwards can be resolved into a component perpendicular to the ramp and a component parallel to the ramp.

The component perpendicular to the ramp is given by:

W_perp = W * cos(theta) = 40lb * cos(15°) = 38.6lb

The component parallel to the ramp is given by:

W_parallel = W * sin(theta) = 40lb * sin(15°) = 10.4lb

where W is the weight of the cart, and theta is the angle of inclination of the ramp.

Next, we resolve the force exerted by the rope into its components. The force exerted by the rope can be resolved into a component perpendicular to the ramp and a component parallel to the ramp.

The component perpendicular to the ramp is given by:

F_perp = F * cos(phi) = F * cos(60°) = 0.5F

The component parallel to the ramp is given by:

F_parallel = F * sin(phi) = F * sin(60°) = 0.87F

where F is the force exerted by the rope, and phi is the angle of inclination of the rope.

To keep the cart from rolling down the ramp, the force exerted by the rope must balance the weight of the cart along the direction of the ramp. That is,

F_parallel = W_parallel

0.87F = 10.4lb

Solving for F, we get:

F = 11.97lb

Therefore, the force that the rope must exert on the cart to keep it from rolling down the ramp is approximately 11.97lb.

User OMGKurtNilsen
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